Calculation of thin-walled vessels. Calculation of thin-walled shells
In engineering practice, structures such as tanks, water reservoirs, gas tanks, air and gas cylinders, building domes, chemical engineering apparatus, parts of turbine and jet engine housings, etc. are widely used. All these structures, from the point of view of their strength and rigidity calculations, can be classified as thin-walled vessels (shells) (Fig. 13.1, a).
A characteristic feature of most thin-walled vessels is that in shape they represent bodies of rotation, i.e. their surface can be formed by rotating some curve 
around the axis ABOUT-ABOUT. Section of a vessel by a plane containing an axis ABOUT-ABOUT, called meridional section, and sections perpendicular to meridional sections are called district. Circumferential sections, as a rule, have the shape of a cone. The lower part of the vessel shown in Fig. 13.1b 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. A 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 thickness of the shell wall exceeds 10 
.
Let us consider the general case of the action of some 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 us select an element from the shell body with two circumferential and two meridional sections (Fig. 13.1, a). The element experiences tension in mutually perpendicular directions and bends. Bilateral tension of an element corresponds to a uniform distribution of normal stresses across the wall thickness 
and the occurrence of normal forces in the shell wall. A change in the curvature of the element suggests the presence of bending moments in the shell wall. When bending, normal stresses arise in the beam wall, varying along the wall thickness.
Under the action of an axisymmetric load, the influence of bending moments can be neglected, since normal forces are predominant. This occurs when the shape of the shell walls and the load on it are such that a balance between external and internal forces is possible without the appearance of bending moments. The theory for calculating shells, based on the assumption that the normal stresses arising in the shell are constant throughout the thickness and, therefore, there is no bending of the shell, is called momentless theory of shells. The momentless theory works well if the shell does not have sharp transitions and hard pinches and, moreover, is not loaded with concentrated forces and moments. Moreover, 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 throughout the wall thickness.
In the presence of concentrated forces and moments, sharp transitions and pinching, solving the problem becomes much more difficult. In places where the shell is attached and in places of sudden changes in shape, increased stresses arise due to the influence of bending moments. In this case, the so-called moment theory of shell calculation. It should be noted that issues 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 where 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 symmetrical shells using the momentless theory. Derivation of Laplace's equation
Let us consider an axisymmetric thin-walled shell experiencing internal pressure from the weight of the liquid (Fig. 13.1, a). Using two meridional and two circumferential sections, we select an infinitesimal element from the shell wall and consider its equilibrium (Fig. 13.2).
In 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: meridional stress 
And hoop stress
. Based on the momentless theory, we will assume that along the wall thickness the stress 
And 
distributed evenly. In addition, we will refer all dimensions of the shell to the middle surface of its walls.
The middle surface of the shell is a surface of double curvature. Let us denote the radius of curvature of the meridian at the point under consideration 
, the radius of curvature of the middle surface in the circumferential direction is denoted by 
. Forces act along the edges of the element 
And 
. Liquid pressure acts on the inner surface of the selected element 
, whose resultant is equal to 
. Let us project the above forces onto the normal 
to the surface:
Let us depict the projection of the element onto the meridional plane (Fig. 13.3) and, based on this figure, write the first term in expression (a). The second term is written by analogy.
Replacing the sine in (a) with 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 into (b) we find:
.
(13.1)
Expression (13.1) represents the equations of Laplace, named after the French scientist who obtained it at the beginning of the 19th century while studying surface tension in liquids.
Equation (13.1) includes two unknown voltages 
And 
. Meridional stress 
we will find by composing the equilibrium equation for the axis 
forces acting on the cut-off part of the shell (Fig. 12.1, b). The circumferential area of the shell walls is calculated using the formula 
. Voltages 
due to the symmetry of the shell itself and the load relative to the axis 
distributed evenly over the area. Hence,
,
(13.2)
Where 
- the weight of the part of the vessel and liquid lying below the section under consideration; 
fluid pressure, according to Pascal’s law, is equal in all directions and equal 
, Where 
depth of the section under consideration, and 
- weight per unit volume of liquid. If a liquid is stored in a vessel under some excess pressure compared to atmospheric 
, then in this case 
.
Now knowing the tension 
from the Laplace 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, 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 acts on the shell) it is equal to 
, and on the opposite – zero. In thin-walled shells, stress 
And 
always much more 
. This means that the magnitude of the third principal stress can be neglected compared to 
And 
, i.e. consider it equal to 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 appropriate strength theory should be used. For example, using the fourth (energy) theory, we write the strength condition in the form:
Let's consider several examples of calculations of momentless shells.
Example 13.1. A spherical vessel is under the influence of uniform internal gas pressure 
(Fig.13.4). Determine the stresses acting in the wall of the vessel and evaluate the strength of the vessel using the third theory of strength. We neglect the own weight of the walls of the vessel and the weight of the gas.
1. Due to the circular symmetry of the shell and the axisymmetric 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 test 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 influence of uniform internal gas pressure 
(Fig. 13.5). Determine the circumferential and meridional stresses acting in the wall of the vessel and evaluate its strength using the fourth theory of strength. Neglect the self-weight of the vessel walls and the weight of the gas.
1. Meridians in the cylindrical part of the shell are generatrices for which 
. From Laplace’s equation (13.1) we find the circumferential stress:
.
(13.6)
2. Using formula (13.2), we find the meridional stress, assuming 
And 
:
.
(13.7)
3. To assess strength, we accept: 
;
;
. The strength condition according to the fourth theory has the form (13.3). Substituting expressions for circumferential and meridional stresses (a) and (b) into this condition, we obtain
Example 12.3. A cylindrical tank with a conical bottom is under the influence of the weight of the liquid (Fig. 13.6, b). Establish the laws of changes in circumferential and meridional stresses within the conical and cylindrical part of the tank, find the maximum stresses 
And 
and construct diagrams of stress distribution along the height of the tank. Neglect the weight of the tank walls.
1. Find the fluid pressure at depth 
:
. (A)
2. We 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 change in circumferential stresses within the conical part of the tank:
.
(13.9)
For the cylindrical part, where 
the distribution law of circumferential stresses has the form:
.
(13.10)
Diagram 
shown in Fig. 13.6, a. For the conical part, this diagram is parabolic. Its mathematical maximum occurs in the middle of the total height at 
. At 
it has a conditional meaning when 
the maximum stress falls within the conical part and has a real value.
Calculation of thin-walled vessels using the momentless theory
Task 1.
The air pressure in the cylinder of the shock-absorbing strut of the aircraft landing gear in the parked position is equal to p = 20 MPa. Cylinder diameter d =….. mm, wall thickness t =4 mm. Determine the main stresses in the cylinder at rest and after takeoff, when the pressure in the shock absorber is ………………….
Answer: (in the parking lot); (after takeoff).
Task 2.
Water enters water turbine through a pipeline whose outer diameter at the machine building is equal to .... m, and the wall thickness t =25 mm. The machine building is located 200 m below the level of the lake from which water is drawn. Find the greatest voltage in ……………………….
Answer:
Task 3.
Check the strength of the wall …………………………… with a diameter of ….. m, under operating pressure p = 1 MPa, if the wall thickness t =12 mm, [σ]=100 MPa. Apply IV strength hypothesis.
Answer:
Task 4.
The boiler has a cylindrical diameter d =…. m and is under operating pressure p=….. MPa. Select the thickness of the boiler wall at the permissible stress [σ]=100 MPa, using III strength hypothesis. What would be the required thickness when using IV strength hypotheses?
Answer:
Task 5.
Steel spherical shell diameter d =1 m and thickness t =…. mm is loaded with internal pressure p = 4 MPa. Determine………………tension and………………..diameter.
Answer: mm.
Task 6.
Cylindrical vessel with diameter d =0.8 m has a wall thickness t =... mm. Determine the permissible pressure in the vessel based on IV strength hypothesis if [σ]=…… MPa.
Answer: [p ]=1.5 MPa.
Task 7.
Define ………………………….. material of a cylindrical shell, if, when loaded with internal pressure, the deformations in the direction of the sensors amounted to
Answer: ν=0.25.
Task 8.
Thick duralumin pipemm and inner diametermm reinforced with a thick steel jacket tightly placed on itmm. Find the limit ………………………..for a two-layer pipe according to the yield strength and ……………… stress between the layers at this moment, assuming E st = 200 GPa,E d =70 GPa,
Answer:
Task 9.
Conduit diameter d =…. mm during the launch period had a wall thickness t =8 mm. During operation, due to corrosion, the thickness in places……………………... What is the maximum column of water that a pipeline can withstand with a double safety margin, if the yield strength of the pipe material is
Problem 10.
Gas pipeline diameter d =……. mm and wall thickness t = 8 mm crosses the reservoir at a maximum ………………………….., reaching 60 m. During operation, gas is pumped under pressure p = 2.2 MPa, and during the construction of an underwater crossing there is no pressure in the pipe. What are they equal to? highest stress in the pipeline and when do they occur?
Problem 11.
A thin-walled cylindrical vessel has hemispherical bottoms. What should be the ratio between the thicknesses of the cylindrical and spherical parts so that in the transition zone there is no………………….?
Problem 12.
When manufacturing railway tanks, they are tested under pressure p = 0.6 MPa. Determine ………………………… in the cylindrical part and in the bottom of the tank, taking the test pressure as the calculated one. Calculate according to III strength hypotheses.
Problem 13.
Between two concentrically located bronze pipes a liquid flows under pressure p = 6 MPa. The thickness of the outer pipe isAt what thickness of the inner pipeis provided by …………………….. of both pipes? What are the highest voltages in this case?
Problem 14.
Determine ………………………… of the shell material if, when loaded with internal pressure, the deformation in the direction of the sensors was
Problem 15.
Thin-walled spherical vessel with diameter d =1 m and thickness t =1 cm is under internal pressure and external What is the ………………….. of the vessel P t if
Would the following solution be correct:
Problem 16.
A thin-walled pipe with plugged ends is under the influence of internal pressure p and bending moment M. Using III strength hypothesis, explore …………………… stressesfrom the value of M for a given r.
Problem 17.
At what depth are the points with ………………….. meridional and circumferential stresses for the conical vessel shown on the right? Determine the magnitude of these stresses, assuming the specific gravity of the product is equal to γ=…. kN/m 3 .
Problem 18.
The vessel is subjected to gas pressure p = 10 MPa. Find………………………if [σ ]=250 MPa.
Answer: t =30 mm.
Problem 19.
A vertically standing cylindrical tank with a hemispherical bottom is filled to the top with water. Thickness of side walls and bottom t =2 mm. Define ………………………. stresses in the cylindrical and spherical parts of the structure.
Answer:
Problem 20.
A cylindrical reservoir is filled to a depth of H 1 = 6 m with liquid of specific gravityand on top - to a thickness of H 2 = 2 m - with water. Determine …………………….. of the tank at the bottom if [σ ]=60 MPa.
Answer: t =5 mm.
Problem 21.
A small gas holder for lighting gas has wall thickness t =5 mm. Find ……………… of the upper and lower vessels.
Answer:
Problem 22.
The valve float of the testing machine is a closed cylinder made of aluminum alloy with a diameter d =…..mm. The float is subjected to………………………pressure р =23 MPa. Determine the thickness of the float wall using the fourth strength hypothesis, if [σ]=200 MPa.
Answer: t =5 mm.
Problem 23.
Thin-walled spherical vessel with a diameter d =1 m and thickness t =1 cm is under the influence of internal ……………… and external What is the ……………….. of the vessel walls If
Answer: .
Problem 24.
Determine the maximum ………………… and circumferential stresses in a toroidal cylinder if p=…. MPa, t =3 mm, A=0.5 mm; d =0.4 m.
Answer:
Problem 25.
Steel hemispherical vessel of radius R =... m is filled with liquid with a specific gravity γ = 7.5 kN/m 3. Taking …………………………. 2 mm and using III strength hypothesis, determine required thickness vessel walls, if [σ]=80 MPa.
Answer: t =3 mm.
Problem 26.
Determine …………………… the points with the highest meridional and circumferential stresses and calculate these stresses if the wall thickness t =... mm, specific gravity of the liquid γ = 10 kN/m 3.
Answer: at a depth of 2 m; at a depth of 4 m.
Problem 27.
A cylindrical vessel with a conical bottom is filled with liquid with a specific gravity γ = 7 kN/m 3. The wall thickness is constant and equal t =...mm. Define …………………………….. and circumferential stresses.
Answer:
Problem 28.
A cylindrical vessel with a hemispherical bottom is filled with liquid with a specific gravity γ = 10 kN/m 3. The wall thickness is constant and equal t =... mm. Determine the maximum stress in the vessel wall. How many times will this voltage increase if the length………………………………, keeping all other dimensions constant?
Answer: will increase by 1.6 times.
Problem 29.
To store oil with a specific gravity γ = 9.5 kN/m 3, a vessel in the form of a truncated cone with a wall thickness is used t =10 mm. Determine the largest …………………………. stress in the vessel wall.
Answer:
Problem 30.
The thin-walled conical bell is located under a layer of water. Determine …………………………….. and hoop stresses if air pressure on the surface under the bell wall thickness t = 10 mm.
Answer:
Problem 31.
Shell thickness t =20 mm, shaped like an ellipsoid of rotation (Ox – axis of rotation), loaded with internal pressure р=…. MPa. Find ………………….. in longitudinal and cross sections.
Answer:
Problem 32.
Using the third strength hypothesis, check the strength of a vessel shaped like a paraboloid of revolution with a wall thickness t =... mm, if the specific gravity of the liquid is γ = 10 kN/m 3, the permissible stress [σ] = 20 MPa, d = h =5 m. Check strength by height……………………………...
Answer: those. strength is ensured.
Problem 33.
A cylindrical vessel with spherical bottoms is designed to store gas under pressure p =... MPa. Under …………………, will it be possible to store gas in a spherical vessel of the same capacity with the same material and wall thickness? What kind of material savings does this achieve?
Answer: savings will be 36%.
Problem 34.
Cylindrical shell with wall thickness t =5 mm compressed by force F =….. kN. Due to manufacturing inaccuracies, the forming shells received little…………………………. Neglecting the influence of this curvature on the meridional stresses, calculatein the middle of the height of the shell, assuming that the generators are curved along one half-wave of the sinusoid, and f =0.01 l; l= r.
Answer:
Problem 35.
A vertical cylindrical vessel is designed to store liquid volume V And specific gravityγ. The total thickness of the upper and lower bases, assigned for design reasons, is equal toDetermine the most favorable height of the tank H opt, at which the mass of the structure will be minimal.Taking the height of the tank equal to H opt, find ………………………….. parts, assuming [σ]=180 MPa, Δ=9 mm, γ=10 kN/m 3, V = 1000 m 3.
Answer: N opt =9 m, mm.
Problem 36.
Long thin tube thick t =…. mm is placed with a tightness Δ on an absolutely rigid rod of diameter d =…..mm . …………… must be applied to the tube to remove it from the rod if Δ=0.0213 mm; f =0.1; l=10 cm, E=100 GPa, ν=0.35.
Answer: F =10 kN.
Problem 37.
A thin-walled cylindrical vessel with spherical bottoms is subjected from the inside to gas pressure p = 7 MPa. By ……………………………….. diameter E 1 =E 2 =200 GPa.
Answer: N 02 =215 N.
Problem 38.
Among others structural elements Cylinders are used in aviation and rocketry high pressure. They usually have a cylindrical or spherical shape and for them, as for other structural units, it is extremely important to comply with the minimum weight requirement. The design of the shaped cylinder shown in the figure is proposed. The walls of the cylinder consist of several cylindrical sections connected by radial walls. Since the cylindrical walls have a small radius, the stress in them decreases, and it can be hoped that despite the increase in weight due to the radial walls, the total weight of the structure will be less than for an ordinary cylinder having the same volume……………………… …….?
Problem 39.
Determine ……………………… of a thin-walled shell of equal resistance containing liquid of specific gravity γ.
Calculation of thick-walled pipes
Task 1.
What is the pressure (internal or external)……………………. pipes? How many times are the greatest equivalent stresses according to III hypothesis of strength in one case more or less than in another if the pressure values are the same? Will the largest radial displacements be equal in both cases?
Task 2.
The two pipes differ only in size cross section: 1st pipe – A=20 cm, b =30 cm; 2nd pipe – A=10 cm, b =15 cm. Which of the pipes has ……………………… ability?
Task 3.
Thick wall pipe with dimensions A=20 cm and b =40 cm cannot withstand the set pressure. In order to increase the load-bearing capacity, two options are proposed: 1) increase the outer radius by P times b ; 2) reduce the internal radius by P times A. Which option gives ……………………………. at same value P?
Task 4.
Pipe with dimensions A=10 cm and b =20 cm withstands pressure p=….. MPa. How much (in percent) ……………….. is the load-bearing capacity of the pipe if the outer radius is increased by … times?
Task 5.
At the end of the First World War (1918), Germany manufactured an ultra-long-range cannon for shelling Paris from a distance of 115 km. It was steel pipe 34 m long and 40 cm thick at the breech. The gun weighed 7.5 MN. Its 120-kilogram projectiles were a meter long with a diameter of 21 cm. The charge used 150 kg of gunpowder, which developed a pressure of 500 MPa, which ejected the projectile with an initial speed of 2 km/s. What should be the ……………………………. used to make a gun barrel, if not less than one and a half times the safety margin?
Task 2. Hydrostatics
Option 0
A thin-walled vessel consisting of two cylinders with diameters D and d, with its lower open end lowered below the liquid level G in reservoir A and rests on supports C located at a height b above this level. Determine the force perceived by the supports if a vacuum is created in the vessel, causing the liquid F in it to rise to a height (a + b). The mass of the vessel is m. How does a change in diameter d affect this force? The numerical values of these quantities are given in Table 2.0.
Table 2.0
|
Liquid F |
||||||
|
Fresh water |
||||||
|
Diesel fuel |
||||||
|
Oil is heavy |
||||||
|
AMG-10 oil |
||||||
|
Transformer |
||||||
|
Spindle |
||||||
|
Turbinnoe |
||||||
|
Light oil |
Option 1
A cylindrical vessel, having a diameter D and filled with liquid to a height a, hangs without friction on a plunger with a diameter d (Fig. 2.1). Determine the vacuum V that ensures equilibrium of the vessel if its mass with lids is m. How do the diameter of the plunger and the depth of its immersion in the liquid affect the result obtained? Calculate the forces in bolted connections B and C of the vessel. The mass of each cover is 0.2 m. The numerical values of these quantities are given in Table 2.1.
Table 2.1
|
Liquid |
|||||
|
Light oil |
|||||
|
Diesel fuel |
|||||
|
Oil is heavy |
|||||
|
AMG-10 oil |
|||||
|
Transformer |
|||||
|
Spindle |
|||||
|
Turbinnoe |
|||||
|
Industrial 20 |
Option 2
The closed tank is divided into two parts by a flat partition, which at depth h has a square hole with side a, closed with a lid (Fig. 2.2). The pressure above the liquid on the left side of the tank is determined by the reading of the pressure gauge p M, the air pressure on the right side by the reading of the vacuum gauge p V. Determine the magnitude of the hydrostatic pressure force on the cover. The numerical values of these quantities are given in Table 2.2.
Table 2.2
|
Liquid |
|||||
|
Diesel fuel |
|||||
|
Light oil |
|||||
|
Oil is heavy |
|||||
|
AMG-10 oil |
|||||
|
Turbinnoe |
|||||
|
Spindle |
|||||
|
Transformer |
|||||
|
Industrial 12 |
In technology, there are often vessels whose walls perceive the pressure of liquids, gases and granular bodies ( steam boilers, tanks, working chambers of engines, tanks, etc.). If the vessels have the shape of bodies of revolution and their wall thickness is insignificant, and the load is axisymmetric, then determining the stresses arising in their walls under load is very 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 throughout 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 wall of the vessel.
We denote the wall thickness t(Fig. 8.1). Radius of curvature of the vessel surface at a given location and Load on the element - internal pressure , normal to the surface of the element.
Let's 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 evenly distributed throughout the wall thickness.
Let's create a condition for the equilibrium of the element, for which we will project the forces acting on the element onto the direction of the normal pp to the surface of the element. The load projection is equal to 
.
The projection of stress onto the normal direction will be represented by a segment ab, equal 
Projection of force acting on edge 1-4 (and 2-3) ,
equal to 
. Similarly, the projection of the force acting on edge 1-2 (and 4-3) is equal to 
.
By projecting all the forces applied to the selected element onto the normal direction pp, we get
Due to the small size of the element, it can be taken
Taking this into account, from the equilibrium equation we obtain
Considering that d 
And
we have
Reduced by 
and dividing by t, we get
(8.1)
This formula is called Laplace's formula. Let's consider the calculation of two types of vessels that are often found in practice: spherical and cylindrical. In this case, we will limit ourselves to cases of internal gas pressure.
| a) b) |
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 strength theory. The principal stresses have the following values: According to the third strength hypothesis; 
. Substituting 
And 
, we get
(8.3)
i.e., strength testing is carried out as in the case of a uniaxial stress state.
According to the fourth strength hypothesis, 
. Since in this case 
, That
(8.4)
i.e., the same condition as under the third strength hypothesis.
2. Cylindrical vessel. In this case 
(cylinder radius) and 
(radius of curvature of the cylinder generatrix).
From Laplace's equation we obtain 
where
(8.5)
To determine the stress, let’s cut the vessel with a plane perpendicular to its axis and consider the equilibrium condition of one of the parts of the vessel (Fig. 47 b).
Projecting onto the axis of the vessel all the forces acting on the cut-off part, we obtain
(8.6)
Where 
-
the resultant of the gas pressure forces at the bottom of the vessel.
Thus, 
,
where
(8.7)
Note that due to the thin-walledness of the ring, which is a cross-section of a cylinder along which stresses act, its area is calculated as the product of the circumference and the wall thickness. Comparing in a cylindrical vessel, we see that
Goal: to form an understanding of the features of deformation and strength calculations of thin-walled shells and thick-walled cylinders.
Calculation of thin-walled shells
Shell - it is a structural element limited by surfaces located at close distances from each other. A shell is called thin-walled if the condition is satisfied for it p/h> 10, where h- shell thickness; p- the radius of curvature of the middle surface, which is the locus of points equidistant from both surfaces of the shell.
Parts whose shape model takes the shell include car tires, vessels, internal combustion engine liners, load-bearing car bodies, aircraft fuselages, ship hulls, floor domes, etc.
It should be noted that shell structures are optimal in many cases, since a minimum of materials are spent on their production.
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.
Let us select from the middle surface in the vicinity of the point TO., lying on this surface, an infinitesimal element 1122 two meridional planes AST And AST 2 s angle d(p between them and two sections normal to the meridians HO t And 220 2 .
Meridional called a section (or plane) passing through the axis of rotation AC. Normal called a section perpendicular to the meridian Sun.
Rice. 73.
Normal sections for the vessel in question are conical surfaces with vertices 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 meridional section;
p,- radius of curvature of the arc 11 in a normal section.
In general r t And p, are a function of the angle V- angle between the axis AC and normal 0,1 (see Fig. 73).
A peculiarity of the operation of shell structures is that all its points, as a rule, are in a complex stress state and strength theories are used to calculate 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 throughout the thickness of the wall.
This theory applies if:
- 1) the shell is a body of revolution;
- 2) shell wall thickness S very small compared to the radii of 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 us to accept the hypothesis that the stress is constant across the wall thickness in a normal section. Based on this hypothesis, we conclude that the walls of the shell work only in tension or compression, since bending is associated with an uneven distribution of normal stresses across the wall thickness.
Let us establish the position of the main areas, i.e. those areas (planes) in which there are no tangential stresses (m = 0).
It is obvious that any meridional section divides the thin-walled shell into two parts, symmetrical both in geometric and force relationships. Since neighboring particles are deformed equally, there is no shear between the sections of the resulting two parts, which means that there are no tangential stresses in the meridional plane (m = 0). Therefore, it is one of the main platforms.
Due to the law of pairing, there will be no tangential stresses in sections perpendicular to the meridional section. Therefore, the normal section (platform) 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 side 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 plane stress state.
To determine the principal stresses, we select a point in the vicinity TO infinitesimal element 1122 (see Fig. 73). Only normal stresses a„ and o, arise on the edges of the element. The first one a t called meridional, and the second A, - circumferential stress, which are the principal stresses at a given point.
Voltage vector A, directed tangent to the circle obtained from the intersection of the middle surface with a normal section. The voltage vector o„ is directed tangentially to the meridian.
Let us express the main stresses through the load (internal pressure) and the geometric parameters of the shell. To determine a 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 Mr. t sin 9, we get
The second equation is obtained from the equilibrium condition of the shell element (Fig. 74, b). If we project all the forces acting on the element onto the normal and equate the resulting expression to zero, we get
Due to small angles we accept
As a result of the mathematical transformations we get the following equation:
This equation is called Laplace equations and establishes the relationship between meridian and circumferential stresses at any point of a thin-walled shell and internal pressure.
Since the dangerous element of the thin-walled shell is in a plane stressed state, based on the results obtained with t And a h and also based on the 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 maximum pressure is reached, the cylindrical vessel (including all pipelines) collapses along the generatrix.
For spherical vessels (p, = r t = g) application of Laplace's equation gives the following results:
_ R g rg _ rg
o, = o t =-, hence, = a 2 = u„= -,
2 h 2 h 2 h
From the results obtained, it becomes obvious that compared to a cylindrical vessel, a spherical vessel is more optimal design. The maximum pressure in a spherical vessel is twice as high.
Let's look at examples of calculating thin-walled shells.
Example 23. Determine the required thickness of the receiver walls if the internal pressure p- 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 t o, R
- -+-=-. It is necessary to find the wall thickness p.
RT P, h
2. Stressed state of the point IN - flat.
Strength condition: er" =cr 1 -et 3?[
- 3. It is necessary to express And o$ through sg„ And A, in letter form.
- 4. Size A", can be found from the equilibrium condition of the cut-off part of the receiver. Voltage value A, - from the Laplace condition, where r t = co.
- 5. Substitute the found values into the strength condition and express the value through them AND.
- 6. For the spherical part, 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.
