Determination of the center of gravity of plane figures. Methods for determining the coordinates of the center of gravity How to determine the center of gravity of irregularly shaped bodies

Note. The center of gravity of a symmetrical figure is on the axis of symmetry.

The center of gravity of the bar is at mid-height. When solving problems, the following methods are used:

1.method of symmetry: the center of gravity of symmetrical figures is on the axis of symmetry;

2. method of separation: we divide complex sections into several simple parts, the position of the centers of gravity of which is easy to determine;

3. negative area method: cavities (holes) are considered as part of a section with a negative area.

Examples of problem solving

Example 1. Determine the position of the center of gravity of the figure shown in Fig. 8.4.

Solution

We split the figure into three parts:

Similarly, it is determined at C = 4.5 cm.

Example 2. Find the position of the center of gravity of a symmetrical bar truss ADBE(fig. 116), the dimensions of which are as follows: AB = 6m, DE = 3 m and EF = 1m.

Solution

Since the truss is symmetrical, its center of gravity lies on the axis of symmetry DF. With the selected (Fig. 116) coordinate system of the abscissa axes of the center of gravity of the truss

The unknown, therefore, is only the ordinate at C center of gravity of the farm. To determine it, we divide the farm into separate parts (rods). Their lengths are determined from the corresponding triangles.

From ΔAEF we have

From ΔADF we have

The center of gravity of each rod lies in its middle, the coordinates of these centers are easily determined from the drawing (Fig. 116).

The found lengths and ordinates of the centers of gravity of individual parts of the farm are entered into the table and using the formula

define the ordinate with the center of gravity of this flat truss.

Hence the center of gravity WITH the whole farm lies on the axis DF symmetry of the truss at a distance of 1.59 m from the point F.

Example 3. Determine the coordinates of the center of gravity of the compound section. The section consists of a sheet and rolled profiles (Fig. 8.5).

Note. Frames are often welded from different profiles to create the required structure. Thus, metal consumption is reduced and a high strength structure is formed.

The intrinsic geometric characteristics are known for standard rolled sections. They are listed in the relevant standards.

Solution

1. Let's designate the figures with numbers and write out the necessary data from the tables:

1 - channel No. 10 (GOST 8240-89); height h = 100 mm; shelf width b= 46 mm; cross-sectional area A 1= 10.9 cm 2;

2 - I-beam No. 16 (GOST 8239-89); height 160 mm; shelf width 81 mm; cross-sectional area And 2 - 20.2 cm 2;

3 - sheet 5x100; thickness 5 mm; width 100mm; cross-sectional area A 3 = 0.5 10 = 5 cm 2.

2. The coordinates of the centers of gravity of each figure can be determined from the drawing.

The compound section is symmetrical, so the center of gravity is on the axis of symmetry and the coordinate X C = 0.

3. Determination of the center of gravity of a composite section:

Example 4. Determine the coordinates of the center of gravity of the section shown in Fig. eight, a. The section consists of two corners 56x4 and channel No. 18. Check the correctness of determining the position of the center of gravity. Indicate its position in the section.

Solution

1. : two corners 56 x 4 and channel No. 18. Let's designate them 1, 2, 3 (see fig. 8, a).

2. We indicate the centers of gravity each profile, using table. 1 and 4 adj. I, and denote them C 1, C 2, C 3.

3. Choose a coordinate system. Axis at compatible with the axis of symmetry, and the axis X will lead through the centers of gravity of the corners.

4. Determine the coordinates of the center of gravity of the entire section. Since the axis at coincides with the axis of symmetry, then it passes through the center of gravity of the section, therefore x with= 0. Coordinate with defined by the formula

Using the tables in the appendix, we determine the areas of each profile and the coordinates of the centers of gravity:

Coordinates at 1 and at 2 are equal to zero, since the axis X passes through the centers of gravity of the corners. Substitute the obtained values ​​into the formula to determine with:

5. We indicate the center of gravity of the section in Fig. 8, a and denote it by the letter C. Let us show the distance at C = 2.43 cm from the axis X to point C.

Since the corners are symmetrically located, have the same area and coordinates, then A 1 = A 2, y 1 = y 2. Therefore, the formula for determining at C can be simplified:

6. Let's check. For this, the axis X draw along the lower edge of the corner shelf (Fig. 8, b). Axis at leave as in the first solution. Formulas for determining x C and at C do not change:

The areas of the profiles will remain the same, and the coordinates of the centers of gravity of the corners and the channel will change. Let's write them out:

Find the coordinate of the center of gravity:

According to the found coordinates x with and with we draw point C on the drawing. The position of the center of gravity found in two ways is at the same point. Let's check it out. Difference between coordinates with, found in the first and second solutions is: 6.51 - 2.43 = 4.08 cm.

This is equal to the distance between the x-axes for the first and second solutions: 5.6 - 1.52 = 4.08 cm.

Answer: with= 2.43 cm, if the x-axis passes through the centers of gravity of the corners, or with = 6.51 cm if the x-axis runs along the bottom edge of the corner shelf.

Example 5. Determine the coordinates of the center of gravity of the section shown in Fig. 9, a. The section consists of an I-beam No. 24 and a channel No. 24a. Show the position of the center of gravity in the section.

Solution

1.Divide the section into rolled profiles: I-beam and channel. Let's designate them with numbers 1 and 2.

3. We indicate the centers of gravity of each profile C 1 and C 2, using the tables of the appendixes.

4. Select the coordinate system. The x axis is compatible with the symmetry axis, and the y axis is drawn through the center of gravity of the I-beam.

5. Determine the coordinates of the center of gravity of the section. The coordinate y c = 0, since the axis X coincides with the axis of symmetry. The coordinate x with is determined by the formula

According to the table. 3 and 4 adj. I and the section diagram, we define

Substitute numerical values ​​into the formula and get

5. Draw point C (the center of gravity of the section) according to the found values ​​of xc and yc (see Fig. 9, a).

The solution must be checked independently with the position of the axes, as shown in Fig. 9, b. As a result of the solution, we obtain x c = 11.86 cm.The difference between the values ​​of x c for the first and second solutions is 11.86 - 6.11 = 5.75 cm, which is equal to the distance between the y axes for the same solutions b dv / 2 = 5.75 cm.

Answer: x c = 6.11 cm, if the y-axis passes through the center of gravity of the I-beam; x c = 11.86 cm, if the y-axis passes through the left extreme points of the I-beam.

Example 6. The railway crane rests on rails, the distance between which is AB = 1.5 m (Fig. 1.102). The gravity force of the crane trolley G r = 30 kN, the center of gravity of the trolley is at point C, lying on the line KL of intersection of the plane of symmetry of the trolley with the plane of the drawing. The force of gravity of the crane winch Q l = 10 kN is applied at the point D. The force of gravity of the counterweight G „= 20 kN is applied at point E. The force of gravity of the boom G c = 5 kN is applied at point H. The outreach of the crane relative to the line KL is 2 m. Determine the stability coefficient of the crane in an unloaded state and what kind of load F can be lifted with this crane, provided that the stability factor must be at least two.

Solution

1. In an unloaded state, the crane has a risk of overturning when turning around the rail A. Therefore, relative to the point A moment of stability

2. The overturning moment about the point A is created by the gravity of the counterweight, i.e.

3. Hence the coefficient of stability of the crane in an unloaded state

4. When the crane jib is loaded with a load F there is a danger of the crane overturning with a turn near rail B. Therefore, relative to the point V moment of stability

5. Overturning moment relative to the rail V

6. By the condition of the problem, the operation of the crane is allowed with the stability coefficient k B ≥ 2, i.e.

Test questions and tasks

1. Why the forces of attraction to the Earth, acting on the points of the body, can be taken as a system of parallel forces?

2. Write down the formulas for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, formulas for determining the position of the center of gravity of plane sections.

3. Repeat the formulas to determine the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half circle.

4.
What is called the static moment of the square?

5. Calculate the static moment of the given figure about the axis Ox. h= 30 cm; b= 120 cm; With= 10 cm (Figure 8.6).

6. Determine the coordinates of the center of gravity of the shaded figure (Fig. 8.7). Dimensions are in mm.

7. Determine the coordinate at figure 1 of the composite section (Fig. 8.8).

When deciding to use the reference data of the tables of GOST "Hot-rolled steel" (see Appendix 1).

Objective – determine the center of gravity of a complex figure analytically and empirically.

Theoretical substantiation. Material bodies consist of elementary particles, the position of which in space is determined by their coordinates. The forces of attraction of each particle to the Earth can be considered a system of parallel forces, the resultant of these forces is called the force of gravity of the body or the weight of the body. The center of gravity of a body is the point of application of the force of gravity.

The center of gravity is a geometric point that can be located outside the body (for example, a disc with a hole, a hollow ball, etc.). Determination of the center of gravity of thin flat homogeneous plates is of great practical importance. Their thickness can usually be neglected and the center of gravity is assumed to be in the plane. If the coordinate plane xOy is aligned with the plane of the figure, then the position of the center of gravity is determined by two coordinates:

where is the area of ​​a part of the figure, ();

- coordinates of the center of gravity of the parts of the figure, mm (cm).

Section of a figure	A, mm 2	X c, mm	Y c, mm
	bh	b / 2	h / 2
	bh / 2	b / 3	h / 3
	R 2 a
For 2α = π πR 2/2

The order of work.

Draw a complex shape, consisting of 3-4 simple shapes (rectangle, triangle, circle, etc.) at a scale of 1: 1 and put down its dimensions.

Draw the coordinate axes so that they cover the entire figure, break a complex figure into simple parts, determine the area and coordinates of the center of gravity of each simple figure relative to the selected coordinate system.

Calculate the coordinates of the center of gravity of the entire figure analytically. Cut this shape out of thin cardboard or plywood. Drill two holes, the edges of the holes should be smooth, and the diameter of the holes is slightly larger than the diameter of the needle for hanging the figure.

First, hang the figure at one point (hole), draw a line with a pencil that coincides with the plumb line. Repeat the same while hanging the figure at a different point. The center of gravity of the figure, found empirically, must match.

Determine the coordinates of the center of gravity of a thin homogeneous plate analytically. Check empirically

Algorithm for solving

1. Analytical method.

a) Draw a drawing on a scale of 1: 1.

b) Break a complex figure into simple ones

c) Select and draw the coordinate axes (if the figure is symmetric, then - along the axis of symmetry, otherwise - along the outline of the figure)

d) Calculate the area of ​​simple shapes and the whole shape

e) Mark the position of the center of gravity of each simple figure in the drawing

f) Calculate the coordinates of the center of gravity of each figure

(x-axis and y-axis)

g) Calculate the coordinates of the center of gravity of the entire figure by the formula

h) Mark the position of the center of gravity in drawing C (

2. Experienced determination.

Check the correctness of the problem solution empirically. Cut this shape out of thin cardboard or plywood. Drill three holes, the edges of the holes should be smooth, and the diameter of the holes is slightly larger than the diameter of the needle for hanging the figure.

First, hang the figure at one point (hole), draw a line with a pencil that coincides with the plumb line. Repeat the same while hanging the figure at other points. The value of the coordinates of the center of gravity of the figure, found when hanging the figure at two points:. The center of gravity of the figure, found empirically, must match.

3. Conclusion on the position of the center of gravity in analytical and experimental determination.

Exercise

Determine the center of gravity of a flat section analytically and experimentally.

Execution example

Task

Determine the coordinates of the center of gravity of a thin homogeneous plate.

I Analytical method

1. The drawing is drawn to scale (dimensions are usually given in mm)

2. Divide a complex figure into simple ones.

1- Rectangle

2- Triangle (rectangle)

3- Area of ​​a semicircle (it is not there, minus sign).

We find the position of the center of gravity of simple shapes of points, and

3. We draw the coordinate axes as convenient and mark the origin of the coordinates.

4. We calculate the area of ​​simple figures and the area of ​​the whole figure. [size in cm]

(3. no, sign -).

Area of ​​the whole figure

5. Find the coordinate of the center. , and in the drawing.

6. Calculate the coordinates of points C 1, C 2 and C 3

7. Calculate the coordinates of point C

8. On the drawing, mark the point

II Empirically

The coordinates of the center of gravity are empirically.

Control questions.

1. Is it possible to consider the force of gravity of a body as a resultant system of parallel forces?

2. Can the center of gravity of the whole body be located?

3. What is the essence of the experimental determination of the center of gravity of a flat figure?

4. How is the center of gravity of a complex figure consisting of several simple figures determined?

5. How should one divide rationally a figure of a complex shape into simple figures when determining the center of gravity of the whole figure?

6. What is the sign of the area of ​​the holes in the formula for determining the center of gravity?

7. At the intersection of which lines of the triangle is its center of gravity?

8. If a figure is difficult to break into a small number of simple figures, what method of determining the center of gravity can give the fastest answer?

Practical work No. 6

"Solving complex problems"

Objective: be able to solve complex problems (kinematics, dynamics)

Theoretical justification: Velocity is a kinematic measure of a point's movement, which characterizes the rate at which its position changes. The speed of a point is a vector that characterizes the speed and direction of movement of a point at a given time. When specifying the motion of a point by equations, the projections of the velocity on the axis of Cartesian coordinates are:

The point velocity module is determined by the formula

The direction of the velocity is determined by the direction cosines:

The characteristic of the rate of change of speed is the acceleration a. The acceleration of a point is equal to the time derivative of the velocity vector:

When specifying the motion of a point, the equations of the projection of acceleration on the coordinate axes are:

Acceleration module:

Full acceleration module

The shear acceleration modulus is determined by the formula

Normal acceleration modulus is determined by the formula

where is the radius of curvature of the trajectory at a given point.

The direction of acceleration is determined by the direction cosines

The equation for the rotational motion of a rigid body around a fixed axis has the form

Body angular velocity:

Sometimes the angular velocity is characterized by the number of revolutions per minute and denoted by a letter. The relationship between and has the form

Angular acceleration of the body:

The force equal to the product of the mass of a given point by its acceleration and direction in the direction directly opposite to the acceleration of the point is called the force of inertia.

Power is the work done by force per unit of time.

Basic equation of dynamics for rotational motion

- the moment of inertia of a body relative to the axis of rotation, is the sum of the products of the masses of material points by the square of their distances to this axis

Exercise

A body of mass m with the help of a cable wound on a drum with a diameter d moves up or down an inclined plane with an angle of inclination α. The equation of motion of the body S = f (t), the equation of rotation of the drum, where S is in meters; φ - in radians; t - in seconds. P and ω - respectively power and angular velocity on the drum shaft at the moment of the end of acceleration or the beginning of deceleration. Time t 1 - acceleration time (from rest to a given speed) or deceleration (from a given speed to a stop). The coefficient of sliding friction between the body and the plane is –f. Disregard the frictional losses on the drum as well as the mass of the drum. When solving problems, take g = 10 m / s 2

No. var	α, deg	Law of motion	Movement direction	m, kg	t 1, s	d, m	P, kW	, rad / s	f	Defined magnitudes
		S = 0.8t 2	Down	-	-	0,20	4,0		0,20	m, t 1
		φ = 4t 2	Down		1,0	0,30	-	-	0,16	P, ω
		S = 1.5t-t 2	up	-	-	-	4,5		0,20	m, d
		ω = 15t-15t 2	up	-	-	0,20	3,0	-	0,14	m, ω
		S = 0.5t 2	Down		-	-	1,76		0,20	d, t 1
		S = 1,5t 2	Down	-	0,6	0,24	9,9	-	0,10	m, ω
		S = 0.9t 2	Down		-	0,18	-		0,20	P, t 1
		φ = 10t 2	Down		-	0,20	1,92	-	0,20	P, t 1
		S = t-1.25t 2	up		-	-	-		0,25	P, d
		φ = 8t-20t 2	up		-	0,20	-	-	0,14	P, ω

Execution example

Problem 1(picture 1).

Solution 1. Rectilinear movement (Figure 1, a). A point, moving uniformly, at some point in time received a new law of motion, and after a certain period of time it stopped. Determine all the kinematic characteristics of the point movement for two cases; a) movement along a straight path; b) movement along a curved trajectory of constant radius of curvature r = 100cm

Figure 1 (a).

The law of change in the speed of a point

We find the initial speed of the point from the condition:

We will find the braking time before stopping from the condition:

at, from here.

The law of motion of a point during a period of uniform motion

The distance traveled by the point along the trajectory during the braking period,

The law of change in the tangential acceleration of a point

whence it follows that during the deceleration period, the point moved equally slowly, since the tangential acceleration is negative and constant in value.

The normal acceleration of a point on a straight trajectory is zero, i.e. ...

Solution 2. Curvilinear movement (Figure 1, b).

Figure 1 (b)

In this case, in comparison with the case of rectilinear motion, all kinematic characteristics remain unchanged, with the exception of normal acceleration.

The law of variation of the normal acceleration of a point

Normal acceleration of a point at the initial moment of deceleration

The numbering of the positions of the point on the trajectory adopted in the drawing: 1 - the current position of the point in uniform motion before the start of braking; 2 - the position of the point at the moment of the start of braking; 3 - the current position of the point during the braking period; 4 - the final position of the point.

Objective 2.

The load (Fig. 2, a) is lifted using a drum winch. The diameter of the drum is d = 0.3m, and the law of its rotation.

The drum accelerated until the angular velocity. Determine all the kinematic characteristics of the movement of the drum and load.

Solution... The law of change in the angular velocity of the drum. We find the initial angular velocity from the condition:; hence, the acceleration started from the rest state. The acceleration time is found from the condition:. The drum rotation angle during the acceleration period.

The law of change in the angular acceleration of the drum, it follows that during the acceleration period the drum rotated uniformly.

The kinematic characteristics of the load are equal to the corresponding characteristics of any point of the traction cable, and hence the point A lying on the drum rim (Fig. 2, b). As you know, the linear characteristics of a point of a rotating body are determined through its angular characteristics.

Distance covered by the load during the acceleration period,. Load speed at the end of acceleration.

Acceleration of cargo.

The law of movement of cargo.

The distance, speed and acceleration of the load could be determined in another way, through the found law of movement of the load:

Objective 3. The load, moving evenly up the inclined reference plane, at some point in time received braking in accordance with the new law of motion , where s is in meters and t is in seconds. The mass of the load is m = 100 kg, the coefficient of sliding friction between the load and the plane is f = 0.25. Determine the force F and the power on the traction cable for two points in time: a) uniform movement before the start of braking;

b) the initial moment of braking. When calculating, take g = 10 m /.

Solution. We determine the kinematic characteristics of the movement of the load.

The law of change in the speed of the cargo

Initial speed of the load (at t = 0)

Acceleration of cargo

Since the acceleration is negative, the movement is slowed down.

1. Uniform movement of the load.

To determine the driving force F, we consider the balance of the load, which is acted upon by a system of converging forces: the force on the cable F, the gravity of the load G = mg, the normal reaction of the support surface N and the friction force directed towards the motion of the body. According to the law of friction,. We select the direction of the coordinate axes, as shown in the drawing, and compose two equilibrium equations for the load:

The power on the cable before the start of braking is determined by the well-known formula

Where m / s.

2. Slow motion of the load.

As you know, with an uneven translational motion of a body, the system of forces acting on it in the direction of motion is not balanced. According to the d'Alembert principle (the method of kinetostatics), the body in this case can be considered as being in conditional equilibrium if, to all the forces acting on it, the inertial force is added, the vector of which is directed opposite to the acceleration vector. The acceleration vector in our case is directed opposite to the velocity vector, since the load is moving in a slow motion. We compose two equilibrium equations for the load:

Power on the cable at the moment of braking

Control questions.

1. How to determine the numerical value and direction of the speed of a point at the moment?

2. What are the normal and tangential components of full acceleration?

3. How to go from expressing angular velocity in min -1 to expressing it in rad / s?

4. What is called body weight? What is the unit of measurement of mass

5. At what motion of a material point does the force of inertia arise? What is its numerical value, how is it directed?

6. Formulate the d'Alembert principle

7. Does the force of inertia arise during the uniform curvilinear motion of a material point?

8. What is Torque?

9. How is the relationship between torque and angular velocity expressed for a given transmitted power?

10. The basic equation of dynamics for rotational motion.

Practical work No. 7

"Structural strength analysis"

Objective: determine strength, section dimensions and permissible load

Theoretical substantiation.

Knowing the force factors and the geometric characteristics of the section during tensile (compression) deformation, we can determine the stress by the formulas. And to understand whether our part (shaft, gear, etc.) can withstand the external load. It is necessary to compare this value with the permissible voltage.

So, the static strength equation

On its basis, 3 types of tasks are solved:

1) strength check

2) determining the dimensions of the section

3) determination of the permissible load

So, the equation of static stiffness

On its basis, 3 types of tasks are also solved.

Static tensile (compressive) strength equation

1) The first type - strength test

,

that is, we solve the left side and compare it with the permissible voltage.

2) The second type - determining the dimensions of the section

from the right side cross-sectional area

Cross section circle

hence the diameter d

Section rectangle

Section square

A = a² (mm²)

Semicircle section

Sections of a channel, I-beam, angle, etc.

Area values ​​- from the table, taken in accordance with GOST

3) The third type is the determination of the permissible load;

taken down, whole number

EXERCISE

Task

A) Strength check (verification calculation)

For a given bar, plot the longitudinal forces and check the strength in both sections. For the material of the bar (steel St3), take

Option No.
	12,5	5,3	-			-
		2,3			-	-
		4,2		-	-

B) Section selection (design calculation)

For a given bar, construct a diagram of longitudinal forces and determine the dimensions of the cross-section in both sections. For the material of the bar (steel St3), take

Option No.
	1,9	2,5
	2,8	1,9
	3,2

C) Determination of the permissible longitudinal force

For a given beam, determine the permissible values ​​of the loads and,

plot the longitudinal forces. For the material of the bar (steel St3), accept. When solving the problem, assume that the type of loading is the same on both sections of the beam.

Option No.
	-	-
			-	-
	-	-

An example of a task

Problem 1(picture 1).

Check the strength of a column made of I-profiles of a given size. For the column material (steel St3), take the permissible tensile stresses and when compressed ... In case of overload or significant underload, select I-beams that provide optimum column strength.

Solution.

A given bar has two sections 1, 2. The boundaries of the sections are sections in which external forces are applied. Since the forces loading the beam are located along its central longitudinal axis, then only one internal force factor arises in the cross sections - the longitudinal force, i.e. stretching (compression) of the bar takes place.

To determine the longitudinal force, we use the section method, the section method. Carrying out a mental section within each of the sections, we will discard the lower fixed part of the bar and leave the upper part for consideration. In section 1, the longitudinal force is constant and equal to

The minus sign indicates that the timber is compressed in both areas.

We build a diagram of longitudinal forces. Having drawn the base (zero) line of the diagram parallel to the axis of the bar, we postpone the obtained values ​​perpendicular to it in an arbitrary scale. As you can see, the diagram turned out to be outlined by straight lines parallel to the base one.

We carry out a check of the strength of the timber, i.e. we determine the calculated stress (for each section separately) and compare it with the allowable one. For this we use the compressive strength condition

where area is the geometric characteristic of the cross-sectional strength. We take from the rolled steel table:

for I-beam
for I-beam

Strength test:

The values ​​of the longitudinal forces are taken in absolute value.

The strength of the timber is ensured, however, there is a significant (more than 25%) underload, which is unacceptable due to excessive consumption of material.

From the strength condition, we determine the new dimensions of the I-beam for each of the sections of the bar:
Hence the required area

According to the GOST table, we select I-beam No. 16, for which;

Hence the required area

According to the GOST table, we select I-beam No. 24, for which;

With the selected sizes of I-beams, there is also an underload, but insignificant (less than 5%)

Problem number 2.

For a bar with given cross-sectional dimensions, determine the permissible load values ​​and. For the material of the bar (steel St3), take the permissible tensile stresses and when compressed .

Solution.

The given bar has two sections 1, 2. There is tension (compression) of the bar.

Using the section method, we determine the longitudinal force, expressing it in terms of the required forces and. Carrying out a section within each of the sections, we will discard the left side of the bar and leave the right side for consideration. In section 1, the longitudinal force is constant and equal to

In section 2, the longitudinal force is also constant and equal to

A plus sign indicates that the bar is stretched in both areas.

We build a diagram of longitudinal forces. The plot is outlined by straight lines parallel to the baseline.

From the condition of the tensile strength, we determine the permissible values ​​of the loads and pre-calculating the areas of the given cross-sections:

Control questions.

1. What internal force factors arise in the cross-section of a bar under tension and compression?

2. Record the tensile and compressive strength condition.

3. How are signs of longitudinal force and normal stress assigned?

4. How will the magnitude of the stress change if the cross-sectional area increases by 4 times?

5. Do the tensile strength and compression strength conditions differ?

6. In what units is voltage measured?

7. Which of the mechanical characteristics is chosen as the ultimate stress for ductile and brittle materials?

8. What is the difference between limit and allowable voltage?

Practical work No. 8

"Solving problems to determine the main central moments of inertia of flat geometric figures"

Objective: determine analytically the moments of inertia of flat bodies of complex shape

Theoretical substantiation. The coordinates of the center of gravity of the section can be expressed in terms of the static moment:

where with respect to the Оx axis

with respect to the Oy axis

The static moment of the area of ​​a figure relative to an axis lying in the same plane is equal to the product of the area of ​​the figure by the distance of its center of gravity to this axis. The static moment has a dimension. The static moment can be positive, negative and zero (relative to any central axis).

The axial moment of inertia of a section is the sum of products taken over the entire section or the integral of elementary areas by the squares of their distances to some axis lying in the plane of the section under consideration

Axial moment of inertia is expressed in units of -. Axial moment of inertia - the quantity is always positive and not equal to zero.

The axes passing through the center of gravity of the figure are called central. The moment of inertia about the central axis is called the central moment of inertia.

The moment of inertia about any axis is equal to the center

Before you can find the center of gravity of simple shapes, such as those that are rectangular, round, spherical or cylindrical, as well as square, you need to know where the center of symmetry of a particular shape is. Since in these cases, the center of gravity will coincide with the center of symmetry.

The center of gravity of a homogeneous bar is located in its geometric center. If it is necessary to determine the center of gravity of a circular disc of a homogeneous structure, then first find the point of intersection of the diameters of the circle. She will be the center of gravity of this body. Considering such figures as a ball, a hoop and a uniform rectangular parallelepiped, we can say with confidence that the center of gravity of the hoop will be in the center of the figure, but outside its points, the center of gravity of the ball is the geometric center of the sphere, and in the latter case, the center of gravity is the intersection diagonals of a rectangular parallelepiped.

Center of gravity of heterogeneous bodies

To find the coordinates of the center of gravity, as well as the center of gravity of an inhomogeneous body, it is necessary to figure out on which segment of the given body the point is located at which all the gravity forces acting on the figure, if it is turned over, intersect. In practice, to find such a point, the body is suspended on a thread, gradually changing the points of attachment of the thread to the body. In the case when the body is in equilibrium, the center of gravity of the body will lie on a line that coincides with the line of the thread. Otherwise, gravity sets the body in motion.

Take a pencil and a ruler, draw vertical straight lines that visually coincide with the thread directions (threads fastened at various points on the body). If the shape of the body is complex enough, then draw several lines that will intersect at one point. It will become the center of gravity for the body you were experimenting with.

Center of gravity of the triangle

To find the center of gravity of a triangle, you need to draw a triangle - a figure consisting of three line segments connected to each other at three points. Before finding the center of gravity of the shape, you need to use a ruler to measure the length of one side of the triangle. Put a mark in the middle of the side, then connect the opposite vertex and the middle of the segment with a line called the median. Repeat the same algorithm with the second side of the triangle, and then with the third. The result of your work will be three medians, which intersect at one point, which will be the center of gravity of the triangle.

If you are faced with a task concerning how to find the center of gravity of a body in the form of an equilateral triangle, then it is necessary to draw a height from each vertex using a rectangular ruler. The center of gravity in an equilateral triangle will be located at the intersection of heights, medians and bisectors, since the same segments are simultaneously heights, medians and bisectors.

Coordinates of the center of gravity of the triangle

Before finding the center of gravity of the triangle and its coordinates, let's take a closer look at the figure itself. This is a homogeneous triangular plate with vertices A, B, C and, accordingly, coordinates: for vertices A - x1 and y1; for vertex В - x2 and y2; for vertex С - x3 and y3. When finding the coordinates of the center of gravity, we will not take into account the thickness of the triangular plate. The figure clearly shows that the center of gravity of the triangle is designated by the letter E - to find it, we drew three medians, at the intersection of which we put the point E. It has its own coordinates: xE and yE.

One end of the median, drawn from vertex A to segment B, has coordinates x 1, y 1, (this is point A), and the second coordinates of the median are obtained based on the fact that point D (the second end of the median) is in the middle of segment BC. The ends of this segment have the coordinates we know: B (x 2, y 2) and C (x 3, y 3). The coordinates of the point D are denoted by xD and yD. Based on the following formulas:

x = (X1 + X2) / 2; y = (Y1 + Y2) / 2

Determine the coordinates of the midpoint of the segment. We get the following result:

xd = (X2 + X3) / 2; yd = (Y2 + Y3) / 2;

D * ((X2 + X3) / 2, (Y2 + Y3) / 2).

We know which coordinates are characteristic for the ends of the blood pressure segment. We also know the coordinates of point E, that is, the center of gravity of the triangular plate. We also know that the center of gravity is located in the middle of the BP segment. Now, applying the formulas and the data we know, we can find the coordinates of the center of gravity.

Thus, we can find the coordinates of the center of gravity of the triangle, or rather, the coordinates of the center of gravity of the triangular plate, given that its thickness is unknown to us. They are equal to the arithmetic mean of the homogeneous coordinates of the vertices of the triangular plate.

Draw a diagram of the system and mark the center of gravity on it. If the found center of gravity is outside the object system, you got the wrong answer. You may have measured distances from different reference points. Repeat measurements.

• For example, if children are sitting on a swing, the center of gravity will be somewhere between the children, not to the right or left of the swing. Also, the center of gravity will never coincide with the point where the child is sitting.
• This reasoning is true in two-dimensional space. Draw a square that will fit all the objects in the system. The center of gravity should be inside this square.

Check the math if you get small results. If the reference point is at one end of the system, the small result places the center of gravity near the end of the system. Perhaps this is the correct answer, but in the vast majority of cases, such a result indicates an error. When you calculated the moments, did you multiply the corresponding weights and distances? If, instead of multiplying, you add the weights and distances, you get a much smaller result.

Correct the error if you find multiple centers of gravity. Each system has only one center of gravity. If you've found multiple centers of gravity, chances are you haven't added all the points. The center of gravity is equal to the ratio of the "total" moment to the "total" weight. You don't have to divide “every” moment by “every” weight: this is how you find the position of each object.

Check the starting point if the answer differs by some integer value. In our example, the answer is 3.4 m. Let's say you received an answer of 0.4 m or 1.4 m, or another number ending in ", 4". This is because you did not choose the left end of the board as a reference point, but a point that is located to the right by a whole amount. In fact, your answer is correct, no matter which starting point you choose! Just remember: the origin is always at x = 0. Here's an example:

• In our example, the origin was at the left end of the board, and we found that the center of gravity is 3.4 m from this origin.
• If you choose a point as a reference point, which is located at a distance of 1 m to the right of the left end of the board, you will get the answer 2.4 m.That is, the center of gravity is at a distance of 2.4 m from the new reference point, which, in turn, located at a distance of 1 m from the left end of the board. Thus, the center of gravity is 2.4 + 1 = 3.4 m from the left end of the board. That's the old answer!
• Note: When measuring distance, remember that the distances to the "left" reference point are negative and to the "right" are positive.

Measure distances in straight lines. Suppose there are two children on the swing, but one child is much taller than the other, or one child is hanging under the board instead of sitting on it. Ignore this difference and measure the straight line distances. Measuring distances at angles will give close but not entirely accurate results.

• In the case of a swing board problem, remember that the center of gravity is between the right and left ends of the board. Later, you will learn how to calculate the center of gravity of more complex two-dimensional systems.

In engineering practice, it happens that it becomes necessary to calculate the coordinates of the center of gravity of a complex plane figure, consisting of simple elements for which the location of the center of gravity is known. Such a task is part of the task of determining ...

Geometrical characteristics of composite cross-sections of beams and bars. Often such questions have to be faced by design engineers of punching dies when determining the coordinates of the center of pressure, developers of loading schemes for various vehicles when placing loads, designers of building metal structures when selecting sections of elements and, of course, students when studying the disciplines "Theoretical Mechanics" and "Resistance 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: a circle, a rectangle (including a square), a parallelogram (including a rhombus), a regular polygon.

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

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

Calculation in Excel of the coordinates of the center of gravity of a compound shape.

It is always easier to convey and perceive information by considering an example than to study a question on purely theoretical calculations. Consider the solution to the problem "How to find the center of gravity?" using the example of the composite shape shown in the figure below this text.

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

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

In cells with yellow fill it with auxiliary preliminary calculations .

Count the results in the cells with light yellow filling.

Blue font is initial data .

Black font is intermediate calculation results .

Red font is final calculation results .

We start solving the problem - we start searching for the coordinates of the center of gravity of the section.

Initial data:

1. We write the names of the elementary figures that form the compound section, respectively.

to cell D3: Rectangle

to cell E3: Triangle

into cell F3: Semicircle

2. Using the "Library of elementary figures" presented in this article, we determine the coordinates of the centers of gravity of the elements of a composite section xci and yci in mm relative to arbitrarily chosen axes 0x and 0y and write

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

xc 1 = a 1 /2

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

yc 1 = b 1 /2

into 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

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

yc 3 = y 03 -4* r3 /3/ π

3. Calculate the area of ​​the elements F 1 , F 2 , F3 in mm2, using again 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 - a semicircle - is negative because this is a cutout - an empty space!

Calculation of the coordinates of the center of gravity:

4. 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 compound 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 merged cell D10E10F10: = D4 * D6 + E4 * E6 + F4 * F6 =80955

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

6. And finally, we 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 is solved, the calculation in Excel is done - 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 very simple in order to make it easier to understand the methodology for calculating the center of gravity of a complex section. The method consists in the fact that any complex figure should be divided into simple elements with known locations of the centers of gravity and the final calculations should be made for the entire section.

If the section is made up of rolled profiles - angles and channels, then they do not need to be divided 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 corner and the channel will be 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 symmetric sections).

The location of the coordinate axes, of course, does not affect the position of the center of gravity of the figure! Therefore, choose a coordinate system that makes your calculations easier. If, for example, I turned the coordinate system 45˚ clockwise in our example, then the calculation of 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 your head”.

The calculated Excel file presented below is not a program in this case. Rather, it is a sketch of a calculator, an algorithm that a template follows in each case. compose 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! A complete calculation of all geometric characteristics of arbitrary complex composite sections will be considered in one of the next articles in the heading "". Follow the news on the blog.

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A few words about the glass, the coin and the 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", causing admiring glances of 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 simply fool our minds!

The center of gravity of the forks + 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 balance! If you shake the forks, it becomes immediately obvious that the system is striving to return to its previous stable position! Imagine a pendulum - an attachment point (= the point of support of the coin on the edge of the glass), the rod-axis of the pendulum (= in our case, the axis is virtual, since the mass of the two forks is separated in different directions of space) and the weight at the bottom of the axis (= the center of gravity of the entire system of “forks + 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 happens with our forks and coins)!

Who does not understand, but wants to understand - figure it out yourself. It's very interesting to "reach" yourself! I will add that the same principle of using a stable balance is implemented in the Vanka-stand up toy. Only the center of gravity of this toy is located above the fulcrum, but below the center of the hemisphere of the supporting surface.

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