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In technology, there are often vessels whose walls perceive the pressure of liquids, gases and bulk solids (steam boilers, reservoirs, engine working chambers, tanks, etc.). If the vessels have the shape of bodies of revolution and their wall thickness is insignificant, and the load is axisymmetric, then the determination of the stresses arising in their walls under load is quite simple.

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

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

Let's cut out an element with dimensions and from the vessel wall.

The wall thickness is denoted by t(fig. 8.1). Radii of curvature of the vessel surface at a given location and Element load - internal pressure , normal to the surface of the element.

Replace the interaction of the element with the rest of the vessel internal forces, the intensity of which is equal to and. Since the wall thickness is insignificant, as already noted, these stresses can be considered uniformly distributed over the wall thickness.

Let us compose the condition for the equilibrium of the element, for which we project the forces acting on the element onto the direction of the normal nn to the surface of the element. Load projection is . The projection of stress on the normal direction will be represented by a segment ab, equal Projection of the force acting on the edge 1-4 (and 2-3) , is equal to ... Similarly, the projection of the force acting on the edge 1-2 (and 4-3) is equal to .

By projecting all the forces applied to the selected element onto the normal direction nn, get

Due to the small size of the element, we can take

Taking this into account, from the equilibrium equation we obtain

Considering that d and we have

Reducing by and dividing into t, we get

(8.1)

This formula is called by the Laplace formula. Consider the calculation of two types of vessels that are often found in practice: spherical and cylindrical. In this case, we will restrict ourselves to the cases of the action of the internal gas pressure.

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

(8.2)

Since in this case there is a plane stress state, then to calculate the strength it is necessary to apply one or another theory of strength. The main voltages have following values: According to the third hypothesis of strength; ... Substituting and , we get

(8.3)

that is, the strength is checked as in the case of a uniaxial stress state.

According to the fourth hypothesis of strength,
... Since in this case , then

(8.4)

that is, the same condition as for the third strength hypothesis.

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

From the Laplace equation we obtain where

(8.5)

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

By projecting all the forces acting on the cut-off part onto the axis of the vessel, we obtain

(8.6)

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

Thus, , where

(8.7)

Note that due to the thinness of the ring, which is a section of a cylinder, along which stresses act, its area is calculated as the product of the circumference by the wall thickness. Comparing and in a cylindrical vessel, we see that

Purpose: to form an idea of ​​the peculiarities of deformation and calculation of the strength of thin-walled shells and thick-walled cylinders.

Calculation of thin-walled shells

Shell - it is a structural element bounded by surfaces located at a close distance from each other. A shell is called thin-walled if the condition p / h> 10 where h - shell thickness; R- the radius of curvature of the median surface, which is the locus of points equidistant from both surfaces of the shell.

Parts, the shape of which take the shell, include car tires, vessels, internal combustion engine liners, load-bearing car bodies, aircraft fuselages, ship hulls, domes, etc.

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

A characteristic feature of most thin-walled shells is that in shape they are bodies of revolution, that is, each of their surfaces can be formed by rotating a certain curve (profile) around a fixed axis. Such bodies of revolution are called axisymmetric. In fig. 73 shows a shell, the middle surface of which is obtained by rotating the profile Sun around the axis AC.

Select from the middle surface in the vicinity of the point TO. lying on this surface, the infinitesimal element 1122 two meridian planes ASt and ASt 2 s angle d (p between them and two sections normal to the meridians HO t and 220 2 .

Meridian called a section (or plane) passing through the axis of rotation AC. Normal called the section perpendicular to the meridian Sun.

Rice. 73.

Normal sections for the vessel under consideration are conical surfaces with tops 0 and Oh g, lying on the axis AC.

Let us introduce the following notation:

r t- radius of curvature of the arc 12 in the meridian section;

R,- radius of curvature of the arc 11 in normal section.

In general r t and R, are a function of the angle v- the angle between the axis AS and normal 0,1 (see fig. 73).

A feature of the operation of shell structures is that all of its points, as a rule, are in a complex stress state, and strength theory is used to calculate the shells.

To determine the stresses arising in a thin-walled shell, the so-called momentless theory. According to this theory, it is believed that there are no bending moments among the internal forces. The walls of the shell work only in tension (compression), and the stresses are evenly distributed over the wall thickness.

This theory is applicable if:

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

The combination of these three conditions allows us to accept the hypothesis of the invariability of the stress across the wall thickness in a normal section. Based on this hypothesis, we conclude that the shell walls work only in tension or compression, since bending is associated with an uneven distribution of normal stresses over the wall thickness.

Let us establish the position of the main areas, that is, those areas (planes) in which there are no shear stresses (t = 0).

It is obvious that any meridional section divides the thin-walled shell into two parts, symmetric both in geometric and in force ratio. Since neighboring particles are deformed in the same way, there is no shear between the sections of the two parts obtained, which means that there are no tangential stresses in the meridional plane (m = 0). Consequently, it is one of the main sites.

By virtue of the law of pairing, there will be no tangential stresses in sections perpendicular to the meridional section. Therefore, the normal section (area) is also the main one.

The third main platform is perpendicular to the first two: at the outer point TO(see Fig. 73) it coincides with the lateral surface of the shell, in it r = o = 0, thus, in the third main area, o 3 = 0. Therefore, the material at the point TO experiences a flat stress state.

To determine the principal stresses, select in the vicinity of the point TO infinitesimal element 1122 (see fig. 73). On the edges of the element, only normal stresses a „and о, arise. The first of these and t called meridian, and the second a, - circumferential stress, which are the main stresses at a given point.

Voltage vector a, directed tangentially to the circle obtained from the intersection of the median surface with a normal section. The stress vector o „is directed tangentially to the meridian.

Let us express the principal stresses in terms of the load (internal pressure) and the geometric parameters of the shell. For determining and t and a, two independent equations are needed. The meridional stress o „can be determined from the equilibrium condition of the cut-off part of the shell (Fig. 74, a):

Substituting rt sin 9, we get

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

In view of the small angles, we take

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

This equation is called Laplace equations and establishes the relationship between the meridian and circumferential stresses at any point of the thin-walled shell and the internal pressure.

Since the dangerous element of a thin-walled shell is in a plane stressed state, based on the results obtained with t and a h and also based on dependence

Rice. 74. Fragment of a thin-walled axisymmetric shell: a) loading scheme; b) stresses acting along the edges of the selected shell element

So, according to the third theory of strength: a "1 = & - st b

Thus, for cylindrical vessels of radius G and wall thickness AND we get

based on the equilibrium equation of the cut-off part, a"

therefore, a, a m, = 0.

When the limiting pressure is reached, the cylindrical vessel (including all pipelines) collapses along its generatrix.

For spherical vessels (R, = p t = d) application of the Laplace equation gives the following results:

_ R r r _ pr

o, = o t =-, hence, = a 2 = and „= -,

2 h 2 h 2 h

From the results obtained, it becomes obvious that, in comparison with a cylindrical vessel, a spherical one is a more optimal design. The limiting pressure in a spherical vessel is twice as high.

Let's consider examples of calculating thin-walled shells.

Example 23. Determine the required wall thickness of the receiver, if the internal pressure R- 4 atm = 0.4 MPa; R = 0.5 m; [a] = 100 MPa (Fig. 75).

Rice. 75.

• 1. In the wall of the cylindrical part, meridian and circumferential stresses arise, related by the Laplace equation: a to o, P
• - + - = -. You need to find the wall thickness NS.

Рт Р, h

2. Point stress V - flat.

Strength condition: er "= cr 1 -t 3? [

• 3. It is necessary to express and about $ across s „ and a, in literal form.
• 4. The value a", can be found from the equilibrium condition of the cut-off part of the receiver. Voltage magnitude a, - from the Laplace condition, where p t = with.
• 5. Substitute the found values ​​into the strength condition and express through them the value AND.
• 6. For the spherical part, the wall thickness h is determined similarly, taking into account p „= p, - R.

1. For a cylindrical wall:

Thus, in the cylindrical part of the receiver o,> o t and 2 times.

Thus, h= 2 mm - thickness of the cylindrical part of the receiver.

Thus, h 2 = 1 mm is the thickness of the spherical part of the receiver.

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Problem 1

Determine the difference in levels of piezometers h.

The system is in equilibrium.

The area ratio of the pistons is 3. H= 0.9 m.

Liquid water.

Task 1.3

Determine the level difference h in piezometers at equilibrium of the pistons of the multiplier, if D/d = 5, H= 3.3 m. Build a graph h = f(D/d), if D/d= 1.5 ÷ 5.

Problem 1. 5

Thin-walled vessel consisting of two cylinders with diameters d= 100 mm and D= 500 mm, the lower open end is lowered under the water level in tank A and rests on supports C located at a height b= 0.5 m above this level.

Determine the magnitude of the force perceived by the supports, if a vacuum is created in the vessel, which caused the rise of water in it to a height a + b= 0.7 m. Empty weight of the vessel G= 300 N. How does the change in diameter affect the result? d?

Task 1.7

Determine the absolute air pressure in the vessel, if the reading of the mercury device h= 368 mm, height H= 1 m. The density of mercury ρ mercury = 13600 kg / m 3. Atmosphere pressure p atm = 736 mm Hg. Art.

Task 1.9

Determine the pressure above the piston p 01 if known: forces on pistons P 1 = 210 N, P 2 = 50 N; meter reading p 02 = 245.25 kPa; piston diameters d 1 = 100 mm, d 2 = 50 mm and height difference h= 0.3 m. Ρ RT / ρ = 13.6.

Target 1.16

Determine pressure p in the hydraulic system and the weight of the load G lying on the piston 2 if for its rise to the piston 1 force applied F= 1 kN. Piston diameters: D= 300 mm, d= 80 mm, h= 1 m, ρ = 810 kg / m 3. Build a graph p = f(D), if D varies from 300 to 100 mm.

Task 1.17.

Determine the maximum height H max, to which gasoline can be sucked in by a piston pump if the pressure of its saturated vapor is h n.p. = 200 mm Hg. Art., and atmospheric pressure h a = 700 mm Hg. Art. What is the force along the rod, if H 0 = 1 m, ρ b = 700 kg / m 3; D= 50 mm?

Build a graph F = ƒ( D) when it changes D from 50 mm to 150 mm.

Target 1.18

Determine the diameter D 1 hydraulic cylinder required to lift the valve in case of excess fluid pressure p= 1 MPa, if the diameter of the pipeline D 2 = 1 m and the mass of the moving parts of the device m= 204 kg. When calculating the coefficient of friction of the gate valve in the guiding surfaces, take f= 0.3, the friction force in the cylinder is considered equal to 5% of the weight of the moving parts. The pressure downstream of the valve is equal to atmospheric, the influence of the stem area should be neglected.

Build a dependency graph D 1 = f(p), if p varies from 0.8 to 5 MPa.

Target 1.19

When the hydraulic accumulator is being charged, the pump supplies water to cylinder A, lifting the plunger B together with the load upwards. When the accumulator is discharged, the plunger, sliding down, squeezes the water out of the cylinder into the hydraulic presses by gravity.

1. Determine the water pressure when charging p s (developed by the pump) and discharge p p (obtained by the presses) of the accumulator, if the mass of the plunger together with the load m= 104 t and plunger diameter D= 400 mm.

The plunger is sealed with a lip, the height of which b= 40 mm and the coefficient of friction on the plunger f = 0,1.

Build a graph p s = f(D) and p p = f(D), if D varies in the range from 400 to 100 mm, the mass of the plunger with the load shall be considered unchanged.

Target 1.21

In a sealed feeding vessel A there is molten babbitt (ρ = 8000 kg / m 3). When the vacuum gauge reads p vac = 0.07 MPa filling the casting ladle B stopped. Wherein H= 750 mm. Determine the height of the babbitt level h in the feeding vessel A.

Target 1.23

Determine strength F required to hold the piston at a height h 2 = 2 m above the surface of the water in the well. A column of water rises above the piston h 1 = 3 m. Diameters: piston D= 100 mm, stock d= 30 mm. Disregard the weight of the piston and rod.

Target 1.24

The vessel contains molten lead (ρ = 11 g / cm 3). Determine the pressure force acting on the bottom of the vessel if the height of the lead level h= 500 mm, vessel diameter D= 400 mm, manovacuum meter reading p vac = 30 kPa.

Construct a graph of the dependence of the pressure force on the diameter of the vessel, if D varies from 400 to 1000 mm

Target 1.25

Determine pressure p 1 fluid that must be supplied to the hydraulic cylinder in order to overcome the force directed along the rod F= 1 kN. Diameters: cylinder D= 50 mm, stock d= 25 mm. Tank pressure p 0 = 50 kPa, height H 0 = 5 m. Do not take into account the friction force. The density of the liquid is ρ = 10 3 kg / m 3.

Target 1.28

The system is in balance. D= 100 mm; d= 40 mm; h= 0.5 m.

What force should be applied to pistons A and B if a force acts on piston C P 1 = 0.5 kN? Friction is neglected. Build a dependency graph P 2 from diameter d, which varies from 40 to 90 mm.

Target 1.31

Determine strength F on the spool stem, if the vacuum gauge reading p vac = 60 kPa, gauge pressure p 1 = 1 MPa, height H= 3 m, piston diameters D= 20 mm and d= 15 mm, ρ = 1000 kg / m 3.

Build a graph F = f(D), if D varies from 20 to 160 mm.

Task 1.32

The system of two pistons connected by a rod is in equilibrium. Determine strength F compressing spring. The liquid between the pistons and in the tank is oil with a density of ρ = 870 kg / m 3. Diameters: D= 80 mm; d= 30 mm; height H= 1000 mm; overpressure R 0 = 10 kPa.

Target 1.35

Determine the load P on cover bolts A and B hydraulic cylinder diameter D= 160 mm, if to a plunger with a diameter d= 120 mm applied force F= 20 kN.

Build a dependency graph P = f(d), if d varies from 120 to 50 mm.

Task1.37

The figure shows a structural diagram of a hydraulic lock, the flow area of ​​which opens when fed into the cavity A control fluid flow with pressure p y. Determine at what minimum value p y piston pusher 1 will be able to open the ball valve if known: spring preload 2 F= 50 H; D = 25 mm, d = 15 mm, p 1 = 0.5 MPa, p 2 = 0.2 MPa. Disregard the forces of friction.

Target 1.38

Determine gauge pressure p m, if the force on the piston P= 100 kgf; h 1 = 30 cm; h 2 = 60 cm; piston diameters d 1 = 100 mm; d 2 = 400 mm; d 3 = 200 mm; ρ m / ρ in = 0.9. Define p m.

Target 1.41

Determine the minimum force value F applied to the rod, under the action of which the piston with a diameter D= 80 mm, if the force of the spring pressing the valve against the seat is F 0 = 100 H, and the fluid pressure p 2 = 0.2 MPa. Valve Inlet (Seat) Diameter d 1 = 10 mm. Rod diameter d 2 = 40 mm, fluid pressure in the rod end of the hydraulic cylinder p 1 = 1.0 MPa.

Target 1.42

Determine the amount of preloading of the differential pressure relief valve spring (mm), which ensures that the valve starts to open when p h = 0.8 MPa. Valve diameters: D= 24 mm, d= 18 mm; spring rate with= 6 N / mm. The pressure to the right of the larger and to the left of the small pistons is atmospheric.

Target 1.44

In a hand-operated hydraulic jack (Fig. 27) at the end of the lever 2 made an effort N= 150 N. The diameters of the pressure head 1 and lifting 4 plungers are respectively equal: d= 10 mm and D= 110 mm. Small lever arm with= 25 mm.

Taking into account the total efficiency of the hydraulic jack η = 0.82, determine the length l lever 2 sufficient to lift the load 3 weight 225 kN.

Build a dependency graph l = f(d), if d varies from 10 to 50 mm.

Objective 1.4 5

Determine the height h column of water in a piezometric tube. The water column balances the full piston with D= 0.6 m and d= 0.2 m, having a height H= 0.2 m. Disregard the dead weight of the piston and the friction in the seal.

Build a graph h = f(D) if the diameter D varies from 0.6 to 1 m.

Target 1.51

Determine the piston diameter = 80.0 kg; depth of water in cylinders H= 20 cm, h= 10 cm.

Build dependency P = f(D), if P= (20 ... 80) kg.

Target 1.81

Determine the reading of the two-liquid pressure gauge h 2, if the pressure on the free surface in the tank p 0 abs = 147.15 kPa, water depth in the tank H= 1.5 m, distance to mercury h 1 = 0.5 m, ρ RT / ρ in = 13.6.

Target 2.33

Air is sucked in by the engine from the atmosphere, passes through the air cleaner and then through a pipe with a diameter d 1 = 50 mm is fed to the carburetor. Air density ρ = 1.28 kg / m 3. Determine the vacuum in the throat of the diffuser with a diameter d 2 = 25 mm (section 2-2) with air flow Q= 0.05 m 3 / s. Take the following resistance coefficients: air cleaner ζ 1 = 5; knee ζ 2 = 1; air damper ζ 3 = 0.5 (referred to the speed in the pipe); nozzle ζ 4 = 0.05 (referred to the velocity in the diffuser throat).

Assignment 18

For weighing heavy loads 3 weighing from 20 to 60 tons, a hydrodynamometer is used (Fig. 7). Piston 1 diameter D= 300 mm, rod 2 diameter d= 50 mm.

Ignoring the weight of the piston and rod, plot the pressure reading R manometer 4 depending on weight m cargo 3.

Assignment 23

In fig. 12 shows a diagram of a hydraulic valve with a spool with a diameter d= 20 mm.

Ignoring the friction in the hydraulic valve and the weight of the spool 1, determine the minimum force that the compressed spring 2 must develop to balance the oil pressure in the lower cavity A R= 10 MPa.

Plot spring force versus diameter d, if d varies from 20 to 40 mm.

Assignment 25

In fig. 14 shows a diagram of a directional valve with a flat valve 2 with a diameter d= 20 mm. In the pressure cavity V hydraulic valve, oil pressure is active p= 5 MPa.

Ignoring cavity back pressure A directional valve and the force of a weak spring 3, determine the length l arm of lever 1, sufficient to open flat valve 2 applied to the end of the lever by force F= 50 N if the length of the small arm a= 20 mm.

Build a dependency graph F = f(l).

Target 1.210

In fig. 10 shows a diagram of a plunger pressure switch, in which when the plunger 3 is moved to the left, the pin 2 rises, which switches the electrical contacts 4. The coefficient of stiffness of the spring 1 WITH= 50.26 kN / m. The pressure switch is activated, i.e. switches electrical contacts 4 at an axial deflection of spring 1 equal to 10 mm.

Ignoring friction in the pressure switch, determine the diameter d plunger, if the pressure switch should be triggered when the oil pressure in cavity A (at the exit) R= 10 MPa.

TaskI.27

Hydraulic booster (pressure boosting device) receives overpressure water from the pump p 1 = 0.5 MPa. In this case, the movable cylinder filled with water A with outer diameter D= 200 mm slides on a fixed rolling pin WITH having a diameter d= 50 mm, creating pressure at the output of the multiplier p 2 .

Determine pressure p 2, taking the friction force in the oil seals equal to 10% of the force developed on the cylinder by pressure p 1, and neglecting the pressure in the return line.

The mass of the moving parts of the multiplier m= 204 kg.

Build a dependency graph p 2 = f(D), if D varies from 200 to 500 mm, m, d, p 1 to be considered constant.

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In engineering practice, such structures as cisterns, water tanks, gas tanks, air and gas cylinders, domes of buildings, chemical engineering apparatus, parts of turbine casings and jet engines, etc. are widely used. All these structures from the point of view of their strength and stiffness calculation can be attributed to thin-walled vessels (shells) (Figure 13.1, a).

A characteristic feature of most thin-walled vessels is that in shape they represent bodies of revolution, i.e. their surface can be formed by rotating some curve around the axis O-O... Section of a vessel by a plane containing an axis O-O is called meridional section, and the sections perpendicular to the meridional sections are called district... The circumferential sections are usually cone-shaped. Shown in Figure 13.1b, the lower part of the vessel is separated from the upper by a circumferential section. The surface dividing the thickness of the walls of the vessel in half is called middle surface... The shell is considered to be thin-walled if the ratio of the smallest principal radius of curvature at a given point on the surface to the shell wall thickness exceeds 10
.

Let us consider the general case of the action of an axisymmetric load on the shell, i.e. such a load that does not change in the circumferential direction and can only change along the meridian. Let's select an element from the shell body with two circumferential and two meridional sections (Fig. 13.1, a). The element is stretched in mutually perpendicular directions and is bent. Bilateral tension of an element corresponds to a uniform distribution of normal stresses along the wall thickness and the occurrence of normal forces in the shell wall. A change in the curvature of an element assumes the presence of bending moments in the shell wall. During bending, normal stresses arise in the beam wall, varying along the wall thickness.

Under the action of an axisymmetric load, the effect of bending moments can be neglected, since normal forces are predominant. This is the case when the shape of the walls of the shell and the load on it are such that balance between external and internal forces is possible without the appearance of bending moments. The theory for calculating shells, built on the assumption that the normal stresses arising in the shell are constant in thickness and, therefore, there is no shell bending, is called momentless shell theory... The momentless theory works well if the shell does not have abrupt transitions and rigid pinches and, moreover, is not loaded with concentrated forces and moments. In addition, this theory gives more accurate results the smaller the shell wall thickness, i.e. the closer to the truth the assumption of a uniform distribution of stresses over the wall thickness.

In the presence of concentrated forces and moments, abrupt transitions and pinches, the solution of the problem is greatly complicated. In the places where the shell is fastened and in places of sharp changes in shape, increased stresses arise due to the influence of bending moments. In this case, the so-called moment theory of shell analysis... It should be noted that questions of the general theory of shells go far beyond the strength of materials and are studied in special sections of structural mechanics. In this manual, when calculating thin-walled vessels, a momentless theory is considered for cases when the problem of determining the stresses acting in the meridional and circumferential sections turns out to be statically determinable.

13.2. Determination of stresses in symmetric shells according to the momentless theory. Derivation of the Laplace equation

Consider an axisymmetric thin-walled shell experiencing internal pressure from the weight of the liquid (Figure 13.1, a). With two meridional and two circumferential sections, select an infinitesimal element from the shell wall and consider its equilibrium (Fig. 13.2).

In the meridional and circumferential sections, there are no tangential stresses due to the symmetry of the load and the absence of mutual displacements of the sections. Consequently, only the main normal stresses will act on the selected element: the meridional stress
and circumferential stress ... Based on the momentless theory, we will assume that the stresses along the wall thickness are
and distributed evenly. In addition, all dimensions of the shell will be referred to the median surface of its walls.

The middle surface of the shell is a surface of double curvature. The radius of curvature of the meridian at the point under consideration is denoted by
, the radius of curvature of the middle surface in the circumferential direction is denoted by ... Forces act on the edges of the element
and
... Liquid pressure acts on the inner surface of the selected element , the resultant of which is
... Project the above forces to the normal
to the surface:

Let us depict the projection of the element onto the meridional plane (Fig. 13.3) and, on the basis of this figure, we write down the first term in expression (a). The second term is written by analogy.

Replacing in (a) the sine by its argument due to the smallness of the angle and dividing all terms of equation (a) by
, we 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)

Expression (13.1) is the Laplace equation, named after the French scientist who received it at the beginning of the 19th century when studying surface tension in liquids.

Equation (13.1) contains two unknown voltages and
... Meridional voltage
we find by composing the equilibrium equation for the axis
forces acting on the cut off part of the shell (Figure 12.1, b). The area of ​​the circumferential section of the walls of the shell is calculated by the formula
... Voltage
due to the symmetry of the shell itself and the load relative to the axis
distributed over the area evenly. Hence,

, (13.2)

where  the weight of the part of the vessel and liquid lying below the considered section;  fluid pressure, according to Pascal's law, is the same in all directions and equal , where Is the depth of the section under consideration, and the weight of a unit volume of liquid. If the liquid is stored in a vessel under some excess in comparison with atmospheric pressure , then in this case
.

Now knowing the tension
from Laplace's equation (13.1) one can find the voltage .

When solving practical problems, due to the fact that the shell is thin, instead of the radii of the middle surface
and substitute the radii of the outer and inner surfaces.

As already noted, the circumferential and meridional stresses and
are the main stresses. As for the third principal stress, the direction of which is normal to the surface of the vessel, then on one of the surfaces of the shell (external or internal, depending on which side the pressure on the shell acts from), it is equal to , and on the opposite - zero. In thin-walled shells of stress and
always much more ... This means that the value of the third principal voltage can be neglected in comparison with and
, i.e. consider it to be zero.

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

Let's consider several examples of calculating momentless shells.

Example 13.1. A spherical vessel is under the action of a uniform internal gas pressure (Figure 13.4). Determine the stresses acting in the vessel wall and evaluate the strength of the vessel using the third theory of strength. We neglect the own weight of the vessel walls and the weight of the gas.

1. Due to the circular symmetry of the shell and the axisymmetry of the stress load and
are the same at all points of the shell. Assuming in (13.1)
,
, a
, we get:

. (13.4)

2. We carry out a check according to the third theory of strength:

.

Considering that
,
,
, the strength condition takes the form:

. (13.5)

Example 13.2. The cylindrical shell is under the action of a uniform internal gas pressure (Figure 13.5). Determine the circumferential and meridional stresses acting in the vessel wall and evaluate its strength using the fourth theory of strength. Neglect the proper weight of the vessel walls and the weight of the gas.

1. Meridians in the cylindrical part of the shell are generators for which
... From the Laplace equation (13.1) we find the circumferential stress:

. (13.6)

2. By the formula (13.2) we find the meridional voltage, assuming
and
:

. (13.7)

3. To assess the strength, we take:
;
;
... The strength condition according to the fourth theory has the form (13.3). Substituting into this condition the expressions for the circumferential and meridional stresses (a) and (b), we obtain

Example 12.3. A cylindrical tank with a conical bottom is under the influence of the weight of the liquid (Figure 13.6, b). Establish the laws of variation of the circumferential and meridional stresses within the conical and cylindrical parts of the reservoir, find the maximum stresses and
and plot stress distribution diagrams along the height of the reservoir. Disregard the weight of the tank walls.

1. Find the fluid pressure at depth
:

... (a)

2. Determine the circumferential stresses from the Laplace equation, taking into account that the radius of curvature of the meridians (generators)
:

... (b)

For the conical part of the shell

;
... (v)

Substituting (c) into (b), we obtain the law of variation of the circumferential stresses within the conical part of the reservoir:

. (13.9)

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

. (13.10)

Diagram shown in Figure 13.6, a. For the conical part, this diagram is parabolic. Its mathematical maximum takes place in the middle of the total height at
... At
it has a conditional meaning, when
the maximum stress falls within the tapered part and has a real value:

. (13.11)

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

... (G)

Substituting (a), (c) and (d) in the formula for meridional stresses (13.2), we get:

. (13.12)

Diagram
shown in Figure 13.6, c. Maximum plot
, outlined for the conical part also along the parabola, takes place at
... It has a real meaning at
when it falls within the tapered part. In this case, the maximum meridional stresses are equal:

. (13.13)

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

. (13.14)

In places where the surface of the reservoir has a sharp break, such as at the transition from the cylindrical to the conical part (Figure 13.7) (Figure 13.5), the radial component of the meridional stresses
not balanced (Figure 13.7).

This component along the perimeter of the ring creates a radial distributed load with an intensity
tending to bend the edges of the cylindrical shell inward. To eliminate this bend, a stiffening rib (spacer ring) is placed in the form of an angle or channel, encircling the shell at the fracture site. This ring takes a radial load (Fig. 13.8, a).

Let's cut out a part of it from the spacer ring with two infinitely closely spaced radial sections (Fig. 13.8, b) and determine the internal forces that arise in it. Due to the symmetry of the spacer ring itself and the load distributed along its contour, lateral force and no bending moment occurs in the ring. Only the longitudinal force remains
... Let's find her.

Let us compose the sum of the projections of all forces acting on the cut out element of the spacer ring onto the axis :

... (a)

Replace the sine of an angle angle due to its smallness
and substitute in (a). We get:

,

(13.15)

Thus, the spacer ring is compressed. The strength condition takes the form:

, (13.16)

where radius of the median line of the ring;  cross-sectional area of ​​the ring.

Sometimes, instead of a spacer ring, a local thickening of the shell is created, bending the edges of the tank bottom inside the shell.

If the shell is under external pressure, then the meridional stresses will be compressive and the radial force becomes negative, i.e. directed outward. Then the stiffening ring will work not in compression, but in tension. In this case, the strength condition (13.16) will remain the same.

It should be noted that the setting of the stiffening ring does not completely eliminate the bending of the shell walls, since the stiffening ring constrains the expansion of the shell rings adjacent to the rib. As a result, the generatrices of the shells near the stiffening ring are bent. This phenomenon is called the edge effect. It can lead to a significant local increase in stresses in the shell wall. The general theory of accounting for the edge effect is considered in special courses using the moment theory of shell calculation.