Task 2. Hydrostatika

Option 0.

The thin-walled vessel consisting of two cylinders with diameters D and D, the lower open end is lowered under the level of fluid in the reservoir A and rests on the supports from the height of the B above this level. Determine the force perceived by the supports if a vacuum was created in the vessel that led to a fluid lifting in it to height (a + b). The mass of the vessel is equal to m. How does the change in diameter D affect this force? The numerical values \u200b\u200bof the specified quantities are shown in Table 2.0.

Table 2.0

	Liquid J.
	Water fresh
	Diesel fuel
	Heavy oil
	AMG-10 oil
	Transform-ane
	Spindle
	Turbid
	Oil is light

Option 1

A cylindrical vessel having a diameter D and filled with a liquid to height A, hanging without friction on the plunger with a diameter D (Fig. 2.1). Determine vacuum V, providing equilibrium of the vessel if its mass with lids M. How affect the result resulting the diameter of the plunger and the depth of its immersion into the liquid? Calculate the forces in bolted connections in and from the vessel. Mass of each lid 0.2 m. The numerical values \u200b\u200bof the specified values \u200b\u200bare shown in Table 2.1.

Table 2.1

	Liquid
	Oil is light
	Diesel fuel
	Heavy oil
	AMG-10 oil
	Transformer
	Spindle
	Turbid
	Industrial 20.

Option 2.

The closed reservoir is divided into two parts with a flat partition, having a square hole with a side A at a depth, closed with a lid (Fig. 2.2). The pressure above the liquid on the left side of the reservoir is determined by the testimony of the pressure gauge R M, the pressure of the air in the right part is the indication of the vacuum meter R v. Determine the value of the hydrostatic pressure for the lid. The numerical values \u200b\u200bof the specified quantities are shown in Table 2.2.

Table 2.2.

					Liquid
					Diesel fuel
					Oil is light
					Heavy oil
					AMG-10 oil
					Turbid
					Spindle
					Transformer
					Industrial 12.

In engineering practice, construction such as tanks, water tanks, gas poles, air and gas cylinders, dome of buildings, chemical engineering devices, part of the hulls of turbines and jet engines, etc. are widely used. All these structures in terms of their calculation on strength and rigidity can be attributed to thin-walled vessels (shells) (Fig.13.1, a).

A characteristic feature of most thin-walled vessels is that in shape they represent the bodies of rotation, i.e. Their surface can be formed by the rotation of some curve. around the axis ABOUT-ABOUT. A vessel section of the plane containing the axis ABOUT-ABOUT, called meridional cross sectionand sections perpendicular to the meridional sections are called district. District sections, as a rule, have a cone. The lower part of the vessel is separated from the upper circumference shown in Figure 13.1b. The surface dividing the thickness of the vessel walls in half is called middle Surface. It is believed that the shell is thin-wing, if the ratio of the smallest main radius of curvature at this surface point to the wall thickness of the shell exceeds the number 10
.

Consider a general case of action on a shell of any axisymmetric load, i.e. Such a load that does not change in the circumferential direction and can only change along the meridian. We highlight the shell from the body two circumferential and two meridional sections element (Fig. 13.1, a). The element is tensile in mutually perpendicular directions and is twisted. Bilateral stretching of the element corresponds to the uniform distribution of normal stresses in the thickness of the wall and the emergence in the wall of the shell of normal effort. The change in the curvature of the element involves the presence of bending moments in the wall. When bending in the beam wall, normal voltages occur, changing through the wall thickness.

Under the action of axisymmetric load, the influence of bending moments can be neglected, since the predominant value is normal forces. This takes place when the shape of the wall of the shell and the load on it is such that it is possible that there is a balance between external and internal efforts without the appearance of bending moments. The theory of calculation of the shells, built on the assumption that the normal stresses arising in the shell are constant in thickness and, therefore, the bend of the shell is missing, called for a reasonable theory of shells. A reasonable theory works well if the shell does not have sharp transitions and rigid pincures and, moreover, not loaded by concentrated forces and moments. In addition, this theory gives more accurate results, the smaller the wall thickness of the shell, i.e. The closer to the truth, the assumption of the uniform distribution of stresses in the thickness of the wall.

In the presence of concentrated forces and moments, sharp transitions and pinches are greatly complicated by the solution of the problem. In places of fastening the shell and in places of sudden changes of the form, elevated voltages arise, due to the influence of bending moments. In this case, the so-called is applied moment theory of calculation of shells. It should be noted that the issues of the general theory of shells go far beyond the resistance of the materials and is studied in special sections of construction mechanics. In this manual, when calculating thin-walled vessels, a reasonable theory is considered for cases when the problem of determining the stresses acting in the meridional and circumferential sections is statically determined.

13.2. Determination of stresses in symmetric shells for a reasonable theory. The output of the Laplas equation

Consider an axisymmetric thin-walled shell, experiencing an internal pressure on the weight of the liquid (Fig.13.1, a). Two meridional and two circumferential sections, select an infinitely small element from the wall of the shell and consider its balance (Fig.13.2).

In meridional and circumferential sections, tangent stresses are absent due to the symmetry of the load and the charter of mutual shifts of sections. Consequently, only the main normal stresses will be valid for the dedicated element: Meridional voltage
and districted voltage . Based on the reasonable theory, we assume that the thickness of the voltage wall
and distributed evenly. In addition, all the sizes of the shell are attributed to the middle surface of its walls.

The median surface of the shell is a surface of a twofold curvature. The radius of the curvature of the meridian in the considered point we denote
, the radius of the curvature of the median surface in the circumferential direction will indicate . Forces on the elements act
and
. The fluid pressure is applied to the inner surface of the dedicated element that is equal to which
. We design the above forces to normal
to the surface:

I will depict the projection of the element per meridional plane (Fig. 13.3) and on the basis of this pattern, we will write down in the expression (a) the first term. The second term is written by analogy.

Replacing in (a) sinus its argument due to the smallness of the angle and delivering all the members of the equation (a) to
We will get:

(b).

Considering that the curvatures of the meridional and circumferential sections of the element are equal, respectively
and
And substituting these expressions in (b) we find:

. (13.1)

The expression (13.1) is the Laplace equations called so in honor of the French scientist who received it at the beginning of the XIXVEK in the study of surface tension in fluids.

The equation (13.1) includes two unknown stresses and
. Meridional tension
find, making the equilibrium equation to the axis
forces acting on the cut-off part of the shell (Fig.12.1, b). The area of \u200b\u200bthe circumferential cross section of the shell walls is considered by the formula
. Voltage
in view of the symmetry of the shell itself and the load relative to the axis
distributed in the area evenly. Hence,

, (13.2)

where ves parts of the vessel and liquids underlying the section under consideration; The liquid's pressure, by the law of Pascal the same in all directions and equal where Glubin of the section under consideration, and ves a unit of liquid volume. If the liquid is stored in a vessel under some excessive comparison with atmospheric pressure , then in this case
.

Now, knowing the voltage
from the Laplace equation (13.1) you can find a voltage .

When solving practical problems in view of the fact that the shell is thin, it is possible instead of radii of the median surface
and enter the radii of the outer and inner surfaces.

As already noted district and meridional stresses and
are main stresses. As for the third main voltage, the direction of which is normal to the surface of the vessel, on one of the surfaces of the shell (external or internal dependence on how the pressure on the shell is applied) it is equal , and on the opposite - zero. In thin-walled stress shells and
always much more . This means that the magnitude of the third main voltage can be neglected compared to and
. Read it equal to zero.

Thus, we assume that the shell material is in a flat intense state. In this case, the appropriate strength theory should be used to assess the strength depending on the state of the material. For example, applying the fourth (energy) theory, the condition of the strength to write in the form:

Consider several examples of calculating the geneless shells.

Example 13.1.Spherical vessel is under the action of uniform internal gas pressure (Fig.13.4). Determine the voltages acting in the vessel wall and estimate the strength of the vessel using the third theory of strength. Own weight of the walls of the vessel and gas weighing neglect.

1. Due to the circular symmetry of the shell and axisymmetry of stress load and
the same in all points of the shell. Believing in (13.1)
,
, but
We get:

. (13.4)

2. Take a check on the third theory of strength:

.

Considering that
,
,
, strength condition take view:

. (13.5)

Example 13.2.The cylindrical shell is under the action of uniform internal gas pressure (Fig.13.5). Determine the district and meridional stresses acting in the vessel wall, and evaluate its strength using the fourth strength theory. Own weight of the walls of the vessel and gas weight neglected.

1. Meridians in the cylindrical part of the shell are forming for which
. From the Laplace equation (13.1) we find the circular voltage:

. (13.6)

2. By formula (13.2) we find a meridional stress, believing
and
:

. (13.7)

3. To assess the strength, we accept:
;
;
. The condition of strength in the fourth theory has the form (13.3). Substituting an expression for circumferential and meridional stresses (a) and (b), we get

Example 12.3.The cylindrical tank with a conical bottom is under the influence of fluid weight (Fig. 13.6, b). Establish the laws of changes in circumferential and meridional stresses within the conical and cylindrical part of the tank, find maximum voltages and
and build voltage distribution plots in the height of the tank. Weight of the walls of the reservoir neglected.

1. Find fluid pressure at depth
:

. (but)

2. Determine the circle voltages from the Laplace equation, given that the radius of the curvature of the meridians (forming)
:

. (b)

For the conical part of the shell

;
. (in)

Substituting (c) in (b) we obtain the law of changes in circumferential stresses within the conical part of the tank:

. (13.9)

For a cylindrical part where
the law of distribution of circumferential stresses has the form:

. (13.10)

Epura. shown in Fig.13.6, a. For the conical part, this parabolic espy. Its mathematical maximum takes place in the middle of the total height when
. For
it has a conditional value when
the voltage maximum falls within the conical part and is of the real value:

. (13.11)

3. Determine the meridional stresses
. For the conical part of the weight of the fluid in the volume of the cone height equal to:

. (d)

Substituting (a), (c) and (g) in a formula for meridional stresses (13.2), we obtain:

. (13.12)

Epura.
shown in Fig.13.6, c. Maximum Epura
defined for the conical part also on Parabola, takes place when
. It has the real value when
When it falls into the limits of the conical part. Maximum meridional voltages are equal to:

. (13.13)

In the cylindrical voltage
in height does not change and equal to the voltage at the upper edge at the suspension place of the tank:

. (13.14)

In places where the surface of the tank has a sharp break, as, for example, at the place of transition from the cylindrical part to conical (Fig.13.7) (Fig.13.5), the radial component of the meridional voltages
not balanced (Fig.13.7).

This component around the perimeter of the ring creates a radial distributed load intensity
, striving to bend the edges of the cylindrical shell inside. To eliminate this bend, the rib of stiffness (spacer ring) is placed in the form of an angle or a chapellery, a shell in a fracture site. This ring perceives the radial load (Fig.13.8, a).

I cut out two infinitely closely located radial sections from the spacer ring, part (Fig.13.8, b) and determine the internal efforts that there are in it. By virtue of the symmetry of the most spacer ring and load distributed by its contour, the transverse force and the bending moment in the ring do not occur. Only longitudinal power remains
. We find it.

We will make up the amount of projections of all forces acting on the cut-out element of the spacer ring, on the axis :

. (but)

Replace sine corner an angle due to his smallness
and we substitute in (a). We get:

,

(13.15)

Thus, the spacer ring works on compression. The condition of strength takes the form:

, (13.16)

where radius midline rings;  Distribution of the cross section of the ring.

Sometimes instead of the spacer ring create a local shell thickening, bending the edges of the bottom of the reservoir inside the shell.

If the shell is experiencing external pressure, meridional voltages will compress and radial force will be negative, i.e. Directed outwards. Then the ring of rigidity will not work on compression, but for stretching. In this case, the condition of strength (13.16) will remain the same.

It should be noted that the formulation of the rings of stiffness does not completely eliminate the bending of the walls of the shell, since the rigid ring is constrained by the expansion of the shell rings adjacent to the edge. As a result, the forming shells near the rigid rings are twisted. The phenomenon is called the edge effect. It can lead to a significant local increase in stresses in the wall of the shell. The general theory of incorporating the regional effect is considered in special courses using the moment theory of calculating the membranes.

The technique often encounters vessels, the walls of which perceive the pressure of liquids, gases and bulk bodies (steam boilers, tanks, engine operating chambers, tanks, etc.). If the vessels have the form of rotation bodies and the thickness of the walls is insignificant, and the load is axisymmetric, then the determination of the voltages arising in their walls under load is produced very simple.

In such cases, without a large error, it can be assumed that only normal stresses (stretching or compressive) arise in the walls and that these stresses are distributed evenly through the thickness of the wall.

Calculations based on such assumptions are well confirmed by experiments, if the wall thickness does not exceed an approximately minimal radius of the curvature of the wall.

I cut the element with dimensions from the wall of the vessel.

The wall thickness is denoted t. (Fig. 8.1). The radii of the curvature of the vessel surface in this place and the load on the element - the internal pressure , Normal to the surface of the element.

We replace the interaction of the element with the remaining part of the vessel inner forces, the intensity of which is equal to and. Since the wall thickness is insignificant, as already noted, these voltages can be considered evenly distributed over the thickness of the wall.

We will make a condition for the equilibrium of the element, for which we will spread the forces acting on the element to the direction of normal pPto the surface of the element. Load projection is equal . The projection of the voltage to the direction of normal will be submitted by the segment aB, equal Projection of the effort acting on the verge of 1-4 (and 2-3) , equal . Similarly, the projection of the effort acting on the verge of 1-2 (and 4-3) is equal to .

Sprogating all the forces attached to the dedicated element, to the direction of normal pp Receive

In view of the smallness of the size of the element can be taken

Given this from the equilibrium equation

Considering that D and have

Reduced by and dividing on t., get

(8.1)

This formula is called laplace formula.Consider the calculation of two types of vessels that are often found in practice: spherical and cylindrical. At the same time, we limit ourselves to the cases of the internal gas pressure.

a) b)

1. Spherical vessel. In this case and From (8.1) follows From

(8.2)

Since in this case there is a flat intense state, it is necessary to apply one or another theory of strength to calculate the strength. The main stresses have the following values: on the third hypothesis of strength; . Substituting and Receive

(8.3)

i.e. verification of strength is carried out, as in the case of a uniaxial intense state.

In the fourth hypothesis of strength,
. As in this case T.

(8.4)

i.e. the same condition as on the third hypothesis of strength.

2. Cylindrical vessel.In this case (cylinder radius) and (radius of curvature forming the cylinder).

From the Laplace equation we get From

(8.5)

To determine the voltage disseminate the vessel with the plane perpendicular to its axis, and consider the equilibrium condition of one of the parts of the vessel (Fig. 47 b).

Projecting on the axis of the vessel all the forces acting on the clipped part, we get

(8.6)

where - remaining gas pressure forces on the bottom of the vessel.

In this way, , From

(8.7)

Note that due to the thin-trialness of the ring, which is a cross section of the cylinder, according to which voltages act, its area is calculated as a product of the circumference of the wall thickness. Comparing both the cylindrical vessel, we see that

If the thickness of the walls of the cylinder is small compared to radii and, then the well-known expression for tangentance stresses acquires the view

i.e. the magnitude determined by us before (§ 34).

For thin-walled reservoirs having the shape of the surfaces of rotation and internal pressure r, distributed symmetrically relative to the axis of rotation, one can derive a general formula for calculating stresses.

We highlight (Fig. 1) from the element under consideration of the element in two adjacent meridional cross sections and two cross sections, normal to the meridian.

Fig.1. Fragment of a thin-walled tank and its intense state.

The dimensions of the element according to the meridian and the direction perpendicular to it will be denoted, respectively, and the radii of the curvature of the meridian and perpendicular to it are denoted and, the wall thickness we call t.

According to symmetry, only normal stresses in the meridial direction and in the direction perpendicular to the meridian will be applied. The corresponding efforts attached to the element faces will be. Since the thin shell resists only with stretching, like a flexible thread, these efforts will be directed by tangential to the meridian and to the cross section, normal to the meridian.

Efforts (Fig. 2) will give in normal to the surface of the element of the direction of the resultant aBequal

Fig.2. Equilibrium element of a thin-walled tank

Similarly, the efforts will be given in the same direction the resultant sum of these efforts balancing the normal pressure applied to the element

This is the main equation that binds the voltage and for thin-walled vessels of rotation is given by Laplas.

Since we set out the distribution (uniform) stresses in the thickness of the wall, the task is statically determined; The second equilibrium equation will turn out if we consider the balance of the lower, cut off by any parallel circle, part of the tank.

Consider the case of hydrostatic load (Fig. 3). Meridional curve Take the axes h. and w. With the start of coordinates at the top of the curve. We spend the section at the level w. From the point ABOUT. The radius of the corresponding parallel circle will be h..

Fig.3. Equilibrium of the lower fragment of a thin-walled tank.

Each pair of efforts acting on the diametrically opposite elements of the section carried out gives a vertical equal bS.equal

the sum of these efforts acting throughout the circumference of the section carried out will be equal to; It will balance the fluid pressure at this level plus the weight of the liquid in the cut part of the vessel.

Knowing the equation of a meridional curve, you can find h. And for each value w., and began to be found, and from the Laplace equation and

For example, for a conical tank with an angle at the top filled with volumetric fluid w.to height h., will have.

Purpose: to form an idea of \u200b\u200bthe features of deformation and calculation on the strength of thin-walled shells and thick-walled cylinders.

Calculation of thin-walled shells

Shell - This element of construction, limited to surfaces located close to each other. The shell is called thin-walled, if a condition is performed for it p / h\u003e 10, where h - shell thickness; r- The radius of the curvature of the median surface, which represents the geometric location of the points equal to both surfaces of the shell.

For details, the form of the forms of which is taken by the shell, include automobile tires, vessels, luminosity sleeves, carrier bodies, fuselages of airplanes, vehicles, ceiling dome, etc.

It should be noted that shell structures in many cases are optimal, since their production is spent by minimum of materials.

A characteristic feature of most thin-walled shells is that in shape they are the bodies of rotation, i.e. each of their surface can be formed by the rotation of some curve (profile) around the stationary axis. Such bodies of rotation are called axisymmetric. In fig. 73 shows the shell, the median surface of which is obtained by rotating the profile Sun around the axis AU.

We highlight from the median surface in the vicinity of the point TO.lying on this surface, infinitely small element 1122 Two meridional planes AST and AST 2 S. Angle d (P. between them and two normal to meridians sections Ho T. and 220 2 .

Meridional called a cross section (or plane) passing through the axis of rotation AU. Normal called cross section perpendicular to meridian Sun.

Fig. 73.

Normal sections for the vessel under consideration are conical surfaces with vertices 0 and O g lying on the axis AU.

We introduce the following notation:

r T. - Radius of the curvature of arc 12 in a meridional cross section;

r, - Radius of the curvature of arc 11 in a normal section.

In general r T. and r, are the function of the corner in - angle between the axis AC and Normal 0,1 (See Fig. 73).

The peculiarity of the work of shell structures is that all its points are usually located in a complex stress state and for calculating the shells apply the theory of strength.

To determine the stresses arising in a thin-walled shell, usually use the so-called so-called without a reasonable theory. According to this theory, it is believed that there are no bending moments among domestic efforts. The walls of the shell work only on stretching (compression), and voltages are evenly distributed over the thickness of the wall.

This theory is applicable if:

• 1) The shell is the body of rotation;
• 2) wall shell thickness S. very small compared to the radius of the curvature of the shell;
• 3) Load, gas or hydraulic pressure are distributed polarly symmetrically relative to the axis of rotation of the shell.

The combination of these three conditions allows the hypothesis about the invaluance of the wall thickness in the normal section. Based on this hypothesis, we conclude that the walls of the shell work only on stretching or compression, since the bending is associated with the uneven distribution of normal stresses in the thickness of the wall.

We establish the position of the main sites, i.e., those sites (planes), in which there are no tangent stresses (T \u003d 0).

It is obvious that any meridional cross-section divides a thin-walled shell into two parts, symmetric both in geometric and power ratio. Since the adjacent particles are deformed equally, then there is no shift between the sections of the two parts obtained, it means that there are no tangent stresses in the meridional plane (T \u003d 0). Consequently, it is one of the main sites.

By virtue of the law, the pair will not be tangent stresses and in sections perpendicular to the meridional cross section. Consequently, the normal cross section (platform) is also the main one.

The third main platform is perpendicular to two first: in the outdoor point TO (See Fig. 73) It coincides with the side shell surface, in it r \u003d o \u003d 0, thus, in the third main site O 3 \u003d 0. Therefore, the material at the point TO Test a flat intense state.

To determine the main stresses, we highlight in the vicinity of the point TO Infinitely small element 1122 (See Fig. 73). On the edges of the element only normal voltages A "and Oh ,. The first of them t. called meridional, And the second but, - district voltage, which are the main stresses at this point.

Voltage vector but, Directed on the tangent of the circle derived from the intersection of the median surface with a normal cross section. The voltage vector is directed by tangent to the meridian.

Express the main stresses through the load (internal pressure) and geometric shell parameters. For determining t. and but, We need two independent equations. Meridional voltage O "can be determined from the equilibrium condition of the clipped part of the shell (Fig. 74, but):

Substation mr. T Sin 9, we get

The second equation is obtained from the equilibrium condition of the shell element (Fig. 74, b). If we design all the forces acting on the element, on the normal and equate the resulting expression zero, then we get

In view of small angles accept

As a result of mathematical transformations, we obtain the equation of the following form:

This equation is called laplas equations and establishes the relationship between the meridianal and circle voltages at any point of the thin-walled shell and internal pressure.

Since the dangerous element of the thin-walled shell is in a flat intense state, based on the results obtained with T. and a H. and also based on dependence

Fig. 74. Fragment of a thin-walled axisymmetric shell: but) loading scheme; b) Voltages acting on the edges of the selected shell element

Thus, on the third theory of strength: a "1 \u003d & - st b

Thus, for cylindrical radius vessels g. and wall thickness AND Receive

based on the equilibrium equation of the clipped part, but"

consequently, a, and t, = 0.

Upon reaching the limit pressure, the cylindrical vessel (including all pipelines) is destroyed by forming.

For spherical vessels (R, = r T \u003d d) The use of the Laplace equation gives the following results:

_ PG RG _ rG

oh, \u003d O T \u003d-, hence, \u003d a 2 \u003d and "= -,

2 h 2 H. 2 h.

It becomes obvious from the results obtained, that, compared with the cylindrical vessel, spherical is a more optimal design. The limit pressure in the spherical vessel is twice as much.

Consider examples of calculating thin-walled shells.

Example 23. Determine the required thickness of the receivers walls if the internal pressure r- 4 atm \u003d 0.4 MPa; R \u003d. 0.5 m; [a] \u003d 100 MPa (Fig. 75).

Fig. 75.

• 1. In the wall of the cylindrical part, meridianal and circumferential stresses associated with the Laplace equation occur: and t o, r
• - + - \u003d -. It is necessary to find the wall thickness p.

RT P, H

2. Stressful point point IN - Flat.

Strength condition: eR "\u003d SG 1 -et 3? [

• 3. It is necessary to express and about $ through sG " and but, In letterproof.
• 4. Value but", You can find from the equilibrium condition of the cut-off part of the receiver. Voltage value but, - from the condition of Laplace, where r T \u003d. CO.
• 5. Substitute the found values \u200b\u200bin the condition of strength and express the magnitude through them AND.
• 6. For spherical part of the wall thickness h. determined similarly, taking into account p "\u003d R, - R.

1. For a cylindrical wall:

Thus, in the cylindrical part of the receiver oh,\u003e O T and 2 times.

In this way, h. \u003d 2 mm - the thickness of the cylindrical part of the receiver.

In this way, h 2 \u003d. 1 mm - thickness of the spherical part of the receiver.