CN101338817B - Safe self-reinforcing pressure vessel - Google Patents

Abstract

The present invention relates to a safe autofrettaged pressure vessel for improving the safety and bearing capability of the vessel and overcoming the technical problem such as unsafety which maybe caused by complex or imprecise calculation of the design of the prior art. The essential points of the technical scheme of the safe autofrettaged pressure vessel are as follows: the depth of a plastic area is calculated according to a formula: k<2>lnk2<j> - k<2> - k2<j> + 2=0, in order to ensure that reverse yield cannot be generated; the bearing capability is calculated according to the graphic formula: P/sigma y =(k<2>-1)/k<2> or P/sigma y =2*(k<2>-1)/(3<1/2>*k<2>), in order to ensure Sigma ej <= Sigma y and Sigma ei <= Sigma y, wherein, k is a ratio between external and internal radiuses of the safe autofrettaged pressure vessel; k<j> is a radio between the radius of an elastic-plastic interface and the internal radius; sigma y is a yield strength of the material; p is an internal pressure borne by the vessel; pe is the maximum elastic bearing capability of a non-autofrettaged vessel; Sigma ej is an equivalent stress of a total stress at the elastic-plastic interface; Sigma ei is the equivalent stress of the total stress at the inner wall surface. When k is less than a value determined by a formula of (k<2>/(k<2>-1))*lnk=1, no matter how large the k<j> is, the container cannot generate reverse yield after autofrettaged pressure is released, and at the moment, the maximum bearing capability of the autofrettaged vessel can reach the full yield pressure, namely the formula of p/sigma y =lnk or p/sigma y =(2/(3<1/2>))*lnk.

Description

A kind of safe self-reinforcing pressure vessel

Technical field

The present invention relates to a kind of safe self-reinforcing pressure vessel, belong to fields such as machine science technology, chemical industry.

Background technique

Pressurized container is the special equipment that is widely used in many industrial departments, as departments such as machinery, chemical industry, pharmacy, the energy, material, food, metallurgy, oil, building, Aeronautics and Astronautics, weapons.The self intensification technology is to improve the important of pressurized container bearing capacity and Security thereof and effective means.The self intensification technology of pressurized container is before manipulating it to be carried out pressure treatment (institute's plus-pressure generally surpasses operation pressure), makes the surrender of cylindrical shell internal layer, produces plastic deformation, form the plastic zone, and skin still is an elastic state.Keep the release after a period of time of this pressure.Release rear cylinder body internal layer plasticity part is because of there being residual deformation can not return to original state, original state is tried hard to return in outer elastic region, but being subjected to stopping of internal layer plastic zone residual deformation can not return to original state, therefore bear stretching, form tensile stress, internal layer then produces pressure stress because being subjected to the compression that skin tries hard to restore.So just formed the pre-stressed state of the outer tension of a kind of internal layer pressurized.After pressing in container comes into operation and bears, prestressing force is superimposed with the interior stress that causes of pressing of operation, and the bigger inboard wall stress of stress is reduced, and the less outer wall stress of stress increases to some extent, and stress is tending towards even in the container wall thereby make.Can improve the bearing capacity of pressurized container thus.Self intensification that Here it is.

The key factor of self intensification technology is the plastic zone degree of depth, i.e. determining of the elasticity of container and plastic zone interface radius, or superstrain degree $\mathrm{\&epsiv;}=\frac{r_j-r_i}{r_o-r_i}\&times;100\%=\frac{k_j-1}{k-1}\&times;100\%$ Determine that wherein ε is the superstrain degree, r _i, r _j, r _oBe respectively inside radius, elastoplasticity interface radius and the outer radius of shell; K is the self-reinforcing pressure vessel outer radius and the ratio of inside radius, i.e. k=r _o/ r _ik _jBe self-reinforcing pressure vessel elasticity and the ratio of plastic zone interface radius with inside radius, i.e. k _j=r _j/ r _i(consulting Fig. 1).The superstrain degree not only has influence on the enforcement of self-reinforcing process, and has influence on bearing capacity of self intensification container or the like.For k _jOr r _jOr ε determine that existing technology mainly contains 1) graphical solution; 2) by formula $r_j=\sqrt{r_o{r}_i}$ Rough calculation; 3) some r are promptly supposed in trial and error method _j, depress r in prestressing force after the calculating self intensification is handled and the operation _jThe equivalent stress σ of the total stress (prestressing force and operational stresses induced sum) at place _Ej, ask for making σ _EjMinimum r _j(this method in fact can be by formula $k_j=\frac{r_j}{r_i}=\exp \left(\frac{p}{{\mathrm{\&sigma;}}_y}\right)$ Or $k_j=\frac{r_j}{r_i}=\exp \left(\frac{\sqrt{3}}{2}\frac{p}{{\mathrm{\&sigma;}}_y}\right)$ Calculate, the former is by the 3rd intensity theory, and the latter presses fourth strength theory).These methods or too rough (as the graphical solution and the estimation technique) can not reflect question essence again; Or too loaded down with trivial details (as trial and error method), can not reflect question essence.And can not overcome some disadvantages, as the reverse yielding problem, i.e. removal self intensification may can produce the secondary compression surrender because being subjected to excessive compression by internal layer after handling time institute's applied pressure.This is very disadvantageous.From the viewpoint of safety, economy, self-reinforcing pressure vessel should guarantee not produce reverse yielding, guarantees r again _jThe equivalent stress σ of the total stress at place _EjLess than yield strength σ _y, bearing capacity is improved.Motive force of the present invention that Here it is.

Summary of the invention

The purpose of this invention is to provide a kind of safe self-reinforcing pressure vessel.

The technical solution adopted for the present invention to solve the technical problems is: the plastic zone degree of depth of self-reinforcing pressure vessel is k by formula ²Lnk _j ²-k ²-k _j ²+ 2=0 calculates, and can guarantee that self intensification does not produce reverse yielding, i.e. σ when handling _Ei'/σ _y〉=-1; Bearing capacity by formula $\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{k^2}=2\frac{p_e}{{\mathrm{\&sigma;}}_y}$ (by the 3rd intensity theory) or $\frac{p}{{\mathrm{\&rho;}}_y}=2\frac{k^2-1}{\sqrt{3}{k}^2}=2\frac{p_3}{{\mathrm{\&sigma;}}_y}$ (pressing fourth strength theory) calculates, and can guarantee σ _Ej≤ σ _yAnd σ _Ei≤ σ _yWherein k is the self-reinforcing pressure vessel outer radius and the ratio of inside radius, k _jBe self-reinforcing pressure vessel elasticity and the ratio of plastic zone interface radius with inside radius, σ _Ei' be the equivalent stress of container inner wall face residual stress behind the removal from strengthen pressure, σ _yBe the self-reinforcing pressure vessel YIELD STRENGTH, p is the interior pressure that self-reinforcing pressure vessel bore, p _eBe the maximum flexibility bearing capacity of non-self-reinforcing pressure vessel, σ _EjBe the equivalent stress of container elastic-plastic district interface place total stress, σ _EiEquivalent stress for container inner wall face place total stress.When k less than by formula $\frac{k^2}{k^2-1}\ln k=1$ The value of determining is when promptly k is less than about 2.2184574899167 (comprise by 2.2184574899167 and obtain approximate number by some rule), no matter k _jMuch, behind the removal from strengthen pressure, container can not produce reverse yielding, and this moment, the maximum load-carrying capacity of self-reinforcing pressure vessel can reach full yield pressure, promptly $\frac{p}{{\mathrm{\&sigma;}}_y}=\ln k$ (the 3rd intensity theory) or $\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{2}{\sqrt{3}}\ln k$ (pressing fourth strength theory).

Beneficial effect of the present invention and advantage are: provide the plastic zone depth calculation formula of self-reinforcing pressure vessel safety, i.e. k ²Lnk _j ²-k ²-k _j ²The thickness size (with k reflection) that+2=0, this formula have established container and the plastic zone degree of depth of safety are (with k _jReflection) function relation between has reflected the name of the game, and that has avoided that prior art calculates is rough or loaded down with trivial details; Found the many maximum k values that can not produce reverse yielding deeply of the plastic zone degree of depth, promptly k equals by formula $\frac{k^2}{k^2-1}\ln k=1$ The value of determining, that is k ≈ 2.2184574899167; The formula of self-reinforcing pressure vessel bearing capacity is provided.

Description of drawings

Fig. 1 is the cross section of pressure container cylinder.

Fig. 2 is pressurized container thickness (k) and the safe plastic zone degree of depth (k _j) relation.

Fig. 3 is based on the bearing capacity figure of the 3rd intensity theory.

Fig. 4 is based on the bearing capacity figure of fourth strength theory.

Embodiment

Embodiment 1, can determine the internal diameter r of pressurized container according to technology Calculation _iContainer material decision back according to it with load (the p/ σ that bears _y), by the bearing capacity formula $\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{k^2}$ (the 3rd intensity theory) or $\frac{p}{{\mathrm{\&sigma;}}_y}=2\frac{k^2-1}{\sqrt{3}{k}^2}$ (fourth strength theory) can determine the footpath than k (according to k=r _o/ r _iCan determine external diameter r _o).After k determines, k by formula ²Lnk _j ²-k ²-k _j ²+ 2=0 determines the ratio k of elastic-plastic district interface radius and inside radius _j, press k _j=r _j/ r _iCan calculate safe r immediately _jKey factor r _jJust can carry out self intensification after determining has handled.Equation k ²Lnk _j ²-k ²-k _j ²Finding the solution of+2=0 can 1) by explicit $k=\sqrt{\frac{{k_j}^2-2}{\ln {k_j}^2-1}}$ Find the solution; Or 2) find the solution with Excel software; Or 3) look into Fig. 2 and get; Or 4) data that provide of according to the form below are looked into and are got (meet intermediate value and can use interpolation):

Table 1 k ²Lnk _j ²-k ²-k _j ²+ 2=0 numerical tables

k k _j k k _j k k _j k k _j

46.12228 1.649 6.662588 1.663 3.364833 1.72 2.290026 1.98

21.58601 1.65 6.5 1.663777 3.211949 1.73 2.282389 1.99

20 1.650212 6.455815 1.664 3.087896 1.74 2.27539 2

19 1.650375 6.268851 1.665 3 1.748442?2.268979 2.01

18 1.650565 6.09877 1.666 2.985168 1.75 2.263112 2.02

17 1.650791 6 1.666625 2.898709 1.76 2.257751 2.03

16.20955 1.651 5.943217 1.667 2.824972 1.77 2.252859 2.04

16 1.651061 5.800273 1.668 2.761395 1.78 2.248404 2.05

15 1.651387 5.668352 1.669 2.706071 1.79 2.244356 2.06

14 1.651788 5.546136 1.67 2.657551 1.8 2.240689 2.07

13.54616 1.652 5.5 1.670397 2.614713 1.81 2.237377 2.08

13 1.652286 5.432516 1.671 2.576674 1.82 2.234398 2.09

12 1.652918 5.32655 1.672 2.542726 1.83 2.231732 2.1

11.88665 1.653 5.227437 1.673 2.512296 1.84 2.22936 2.11

11 1.653735 5.134487 1.674 2.5 1.844363?2.227263 2.12

10.72744 1.654 5.047102 1.675 2.484917 1.85 2.225426 2.13

10 1.65482 5 1.675565 2.460201 1.86 2.223835 2.14

9.859732 1.655 4.964763 1.676 2.437823 1.87 2.222475 2.15

9.1793 1.656 4.887016 1.677 2.417512 1.88 2.221333 2.16

9 1.656304 4.813461 1.678 2.399035 1.89 2.220397 2.17

8.627561 1.657 4.743747 1.679 2.382195 1.9 2.219658 2.18

8.168728 1.658 4.67756 1.68 2.366822 1.91 2.219103 2.19

8 1.658414 4.5 1.682956 2.35277 1.92 2.218724 2.2

7.779543 1.659 4.160525 1.69 2.33991 1.93 2.218513 2.21

7.44415 1.66 4 1.694172 2.328131 1.94 2.218459 2.22

7.151321 1.661 3.811702 1.7 2.317334 1.95 2.218457 2.218457

7 1.661571 3.558259 1.71 2.307434 1.96

6.892851 1.662 3.5 1.712755 2.298354 1.97

Annotate: this example focuses on the key point of this patent, so the existing design procedure of pressurized container is not had and needn't be described in detail.

Analyze below in conjunction with accompanying drawing.At first consider situation by the 3rd intensity theory.Shown in Figure 1 is the cross section of a pressure container cylinder, and internal layer is the plastic zone, and skin is the elastic region, and the elastic-plastic interface radius is r _j

According to the existing theory of pressurized container, behind the removal from strengthen pressure, the residual stress in the container wall is:

The plastic zone:

$\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{2}[\frac{k_j^2}{k^2}+\ln \frac{{(r/r_i)}^2}{k_j^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}]---\left(1\right)$

$\frac{{\mathrm{\&sigma;}}_r\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{2}[\frac{k_j^2}{k^2}-1+\ln \frac{{(r/r_i)}^2}{k_j^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}(1-\frac{k^2}{{(r/r_i)}^2})]---\left(2\right)$

$\frac{{\mathrm{\&sigma;}}_t\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{2}[\frac{k_j^2}{k^2}+1+\ln \frac{{(r/r_i)}^2}{k_j^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}(1+\frac{k^2}{{(r/r_i)}^2})]---\left(3\right)$

The elastic region:

$\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{2}[\frac{k_j^2}{k^2}-(1-\frac{k_j^2}{k^2}+{\ln k}_j^2)\frac{1}{k^2-1}]---\left(4\right)$

$\frac{{\mathrm{\&sigma;}}_r\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{2}(1-\frac{k^2}{{(r/r_i)}^2})[\frac{k_j^2}{k^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}]=(1-\frac{k^2}{{(r/r_i)}^2})\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}---\left(5\right)$

$\frac{{\mathrm{\&sigma;}}_t\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{2}(1+\frac{k^2}{{(r/r_i)}^2})[\frac{k_j^2}{k^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}]=(1+\frac{k^2}{{(r/r_i)}^2})\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}---\left(6\right)$

σ wherein _z'---axial residual stress; σ _r'---residual stress radially;

σ _t'---circumferential residual stress; R---the radius at place, arbitrfary point in the container wall.

Can get elastic-plastic (r=r at the interface by formula (1)-(3) or (4)-(6) _j) residual stress:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{2}[\frac{{r_j}^2}{{r_o}^2}-(1-\frac{{r_j}^2}{{r_o}^2}+2\ln \frac{r_j}{r_i})\frac{1}{k^2-1}]=\frac{1}{2}[\frac{{k_j}^2}{k^2}-(1-\frac{{k_j}^2}{k^2}+2\ln k_j)\frac{1}{k^2-1}]---\left(7\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{rj}}\&prime;}{{\mathrm{\&sigma;}}_y}=(1-\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}\&prime;}{{\mathrm{\&sigma;}}_y}---\left(8\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{rj}}\&prime;}{{\mathrm{\&sigma;}}_y}=(1+\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}\&prime;}{{\mathrm{\&sigma;}}_y}---\left(9\right)$

The interior p that presses is at r _jThe stress that the place causes is:

$\begin{array}{ccc}\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^p}{{\mathrm{\&sigma;}}_y}=\frac{1}{k^2-1}\frac{p}{{\mathrm{\&sigma;}}_y}& \frac{{\mathrm{\&sigma;}}_{\mathrm{rj}}^p}{{\mathrm{\&sigma;}}_y}=(1-\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^p}{{\mathrm{\&sigma;}}_y}& \frac{{\mathrm{\&sigma;}}_{\mathrm{tj}}^p}{{\mathrm{\&sigma;}}_y}=(1+\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^p}{{\mathrm{\&sigma;}}_y}---\left(10\right)\end{array}$

r _jThe total stress at place is exactly the addition of formula (7) to the corresponding formula of formula (10):

$\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^T}{{\mathrm{\&sigma;}}_y}=\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}\&prime;}{{\mathrm{\&sigma;}}_y}+\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^p}{{\mathrm{\&sigma;}}_y}---\left(11\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{rj}}^T}{{\mathrm{\&sigma;}}_y}=(1-\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}\&prime;}{{\mathrm{\&sigma;}}_y}+(1-\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^p}{{\mathrm{\&sigma;}}_y}=(1-\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^T}{{\mathrm{\&sigma;}}_y}---\left(12\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{tj}}^T}{{\mathrm{\&sigma;}}_y}=(1+\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}\&prime;}{{\mathrm{\&sigma;}}_y}+(1+\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^p}{{\mathrm{\&sigma;}}_y}=(1+\frac{k^2}{k_j^2})\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^T}{{\mathrm{\&sigma;}}_y}---\left(13\right)$

By the 3rd intensity theory, r _jThe equivalent stress of locating the total stress of three directions (z, r, t direction) is:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ej}}}{{\mathrm{\&sigma;}}_y}=2\frac{k^2}{k_j^2}\frac{{\mathrm{\&sigma;}}_{\mathrm{zj}}^T}{{\mathrm{\&sigma;}}_y}=1-\frac{k^2}{k_j^2(k^2-1)}(1-\frac{k_j^2}{k^2}+2\ln k_j)+\frac{2k^2}{k_j^2(k^2-1)}\frac{p}{{\mathrm{\&sigma;}}_y}---\left(14\right)$

Can get the container inner wall face (r=r of place by formula (1)-(3) _i) residual stress:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}\&prime;}{{\mathrm{\&sigma;}}_y}=-\frac{1}{2(k^2-1)}(1-k_j^2+2k^2\ln k_j)---\left(15\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ti}}\&prime;}{{\mathrm{\&sigma;}}_y}=-\frac{1}{(k^2-1)}(1-k_j^2+2\ln k_j)=2\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}\&prime;}{{\mathrm{\&sigma;}}_y}---\left(16\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ri}}\&prime;}{{\mathrm{\&sigma;}}_y}=0---\left(17\right)$

By the 3rd intensity theory, r _iThe equivalent stress of place's residual stress is:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{{\mathrm{\&sigma;}}_{\mathrm{ti}}\&prime;}{{\mathrm{\&sigma;}}_y}=2\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}\&prime;}{{\mathrm{\&sigma;}}_y}=-\frac{1}{k^2-1}(1-k_j^2+{2k}^2\ln k_j)---\left(18\right)$

σ _Ei'/σ _yMust be pressure stress, and along with k _jIncrease and increase (absolute value increase); During k → ∞, σ _Ei'/σ _y→

(or e ^0.5).

The interior p that presses is at r _iThe stress that the place causes is:

$\begin{array}{ccc}\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}=\frac{1}{k^2-1}\frac{p}{{\mathrm{\&sigma;}}_y}& \frac{{\mathrm{\&sigma;}}_{\mathrm{ri}}^p}{{\mathrm{\&sigma;}}_y}=(1-k^2)\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}& \frac{{\mathrm{\&sigma;}}_{\mathrm{ti}}^p}{{\mathrm{\&sigma;}}_y}=(1+k^2)\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}\end{array}$

Total stress (stress that residual stress+interior pressure p causes) is:

$\begin{array}{ccc}\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^T}{{\mathrm{\&sigma;}}_y}=\frac{{{\mathrm{\&sigma;}}_{\mathrm{zi}}}^{\&prime;}}{{\mathrm{\&sigma;}}_y}+\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}& \frac{{\mathrm{\&sigma;}}_{\mathrm{ri}}^T}{{\mathrm{\&sigma;}}_y}=(1-k^2)\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}& \frac{{\mathrm{\&sigma;}}_{\mathrm{ti}}^T}{{\mathrm{\&sigma;}}_y}=2\frac{{{\mathrm{\&sigma;}}_{\mathrm{zi}}}^{\&prime;}}{{\mathrm{\&sigma;}}_y}(1+k^2)\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}\end{array}$

By the 3rd intensity theory, r _iThe equivalent stress of place's total stress is:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}}{{\mathrm{\&sigma;}}_y}=2\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}\&prime;}{{\mathrm{\&sigma;}}_y}+2k^2\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}=\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;}{{\mathrm{\&sigma;}}_y}+\frac{2k^2}{k^2-1}\frac{p}{{\mathrm{\&sigma;}}_y}---\left(19\right)$

Actual following formula (18) can make p/ σ by formula (19) _y=0 obtains.

Passing through type (18), formula (14) can be write as:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ej}}}{{\mathrm{\&sigma;}}_y}=\frac{{k_j}^2-1}{{k_j}^2}+\frac{2k^2}{{k_j}^2(k^2-1)}\frac{p}{{\mathrm{\&sigma;}}_y}+\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;/{\mathrm{\&sigma;}}_y}{{k_j}^2}---\left(14a\right)$

Make formula (1)=formula (2) and formula (1)=formula (3) all get:

${\left(\frac{r}{r_i}\right)}^2=k_r^2=\frac{1}{k^2-1}(k^2-k_j^2+k^2\ln {k_j}^2)=\left|\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;}{{\mathrm{\&sigma;}}_y}\right|+1$

Three residual stress distribution curves of this explanation meet at a bit.

Make p/ σ in formula (14) or (14a) _y=0 gets r _jThe equivalent stress of the residual stress at place:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ej}}\&prime;}{{\mathrm{\&sigma;}}_y}=1-\frac{1}{k^2-1}(\frac{k^2}{k_j^2}-1+\frac{2k^2}{k_j^2}\ln k_j)=\frac{{k_j}^2-1}{{k_j}^2}+\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;/{\mathrm{\&sigma;}}_y}{{k_j}^2}--\left(20\right)$

σ _Ej'/σ _yMust be tensile stress, and σ _Ei'/σ _yCompare σ _Ej'/σ _yMore dangerous.

In formula (18), make k _j=k,

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;}{{\mathrm{\&sigma;}}_y}=1-\frac{2k^2}{k^2-1}\ln k---\left(21\right)$

In formula (21), make σ _Ei'/σ _y=-1:

$\frac{k^2}{k^2-1}\ln k=1---\left(22\right)$

In other words, when k less than by formula $\frac{k^2}{k^2-1}\ln k=1$ The value of determining, promptly k is less than about 2.2184574899167 o'clock, no matter k _jMuch, behind the removal from strengthen pressure, container can not produce reverse yielding.

In formula (18), directly make σ _Ei'/σ _y=-1:

k ²Lnk _j ²-k ²-k _j ²+ 2=0 or $k=\sqrt{\frac{{k_j}^2-2}{\ln {k_j}^2-1}}---\left(23\right)$

This formula association k (reflection container thickness size) and k _j(the reflection plastic zone degree of depth) shown k and k _jBetween clear and definite function relation.This is very crucial, utilizes this formula, can make things convenient for and easily calculates the safe plastic zone degree of depth (k for the container of any thickness _j).How using formula (23) has provided several different methods in embodiment 1.According to formula (23), from table 1 and Fig. 2 as seen, k is big more, and promptly cylindrical shell is thick more, k _jMore little, promptly the plastic zone degree of depth is shallow more.This uses great convenience to engineering is provided, because bigger yield region is given birth in the thick more difficult labour more of cylindrical shell, so can take more shallow plastic zone when cylindrical shell is thicker, not only can not produce reverse yielding when self intensification is handled, and can meet design requirement.

In formula (14), make σ _Ej/ σ _y=1 must the plastic zone degree of depth be k _jPressurized container work as r _jThe load that the place can bear when just having begun to surrender:

$\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{1}{2k^2}(k^2-{k_j}^2+2k^2\ln k_j)---\left(24\right)$

Formula (23) is write as

2k ²Lnk _j-k _j ²=k ²-2 substitution formulas (24):

$\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{k^2}=2\frac{p_e}{{\mathrm{\&sigma;}}_y}$ k≥2.2184574899167 (25)

The bearing capacity formula of the self-reinforcing pressure vessel of safety that Here it is, its image as shown in Figure 3.Do some explanations below convolution (25) and the image thereof:

(1) curve ofa (k=1～2.2184574899167).When being in this stage in the self intensification processing procedure up to k _jReverse yielding can not take place in=k.The maximum load-carrying capacity of self intensification container can reach its full yield pressure p/ σ _y=lnk (k _j=k).There is σ in this stage _Ei'/σ _y=-1 (is equivalent to k ²Lnk _j ²-k ²-k _j ²+ 2=0), σ _Ej=σ _yσ _EjminWhile σ _Ei=σ _yIn case (because σ _Ej=σ _yσ must be arranged _Ei=σ _y), σ _Ej'/σ _y＜1.

(2) curve ab (k 〉=2.2184574899167).This stage has:

σ _Ei'/σ _y=-1 (is equivalent to k ²Lnk _j ²-k ²-k _j ²+ 2=0), σ _Ej=σ _yσ _Ejmin, while σ _Ei=σ _yBearing capacity is: $\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{k^2}=2\frac{p_e}{{\mathrm{\&sigma;}}_y}$

P/ σ _yIncrease along with the k increase (and k _jReduce along with the k increase); During k → ∞, k _j→ exp (0.5), p/ σ _y→ 1; This stage bearing capacity greater than the od section less than the ac section.

(3) dotted line ac (k 〉=2.2184574899167): p/ σ _y=p _y/ σ _y=lnk.

This section curve and solid line ofa are same, but behind k 〉=2.2184574899167, p/ σ _yAnd can't help ac and determine.

(4) dotted line od (k=1～∞).This is the maximum flexibility bearing capacity of non-self intensification container $\frac{p_e}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{2k^2},$ It is a correlation curve.

(5) dotted line oea (k=1～2.2184574899167).With solid line ab be same $(\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{k^2}),$ Bearing capacity on this section compares p _y/ σ _yAlso big.But in k=1～2.2184574899167 stages, bearing capacity be can't help curve oea and by curve ofa, determined, otherwise σ _EiWith σ _EjAll will be above σ _y, this is unsafe.

K＜2.2184574899167 o'clock, | σ _Ei'/σ _y| permanent in 1.

Therefore, curve ofab is best bearing capacity curve.

Press fourth strength theory.According to the existing theory of pressurized container, have corresponding to formula (1)-(6):

The plastic zone:

$\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{\sqrt{3}}[\frac{k_j^2}{k^2}+\ln \frac{{(r/r_i)}^2}{k_j^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}]---\left(1a\right)$

$\frac{{\mathrm{\&sigma;}}_r\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{\sqrt{3}}[\frac{k_j^2}{k^2}-1+\ln \frac{{(r/r_i)}^2}{k_j^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}(1-\frac{k^2}{{(r/r_i)}^2})]---\left(2a\right)$

$\frac{{\mathrm{\&sigma;}}_r\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{\sqrt{3}}[\frac{k_j^2}{k^2}+1+\ln \frac{{(r/r_i)}^2}{k_j^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}(1+\frac{k^2}{{(r/r_i)}^2})]---\left(3a\right)$

The elastic region:

$\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{\sqrt{3}}[\frac{k_j^2}{k^2}-(1-\frac{k_j^2}{k^2}+{\ln k}_j^2)\frac{1}{k^2-1}]---\left(4a\right)$

$\frac{{\mathrm{\&sigma;}}_r\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{\sqrt{3}}(1-\frac{k^2}{{(r/r_i)}^2})[\frac{k_j^2}{k^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}]=(1-\frac{k^2}{{(r/r_i)}^2})\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}---\left(5a\right)$

$\frac{{\mathrm{\&sigma;}}_t\&prime;}{{\mathrm{\&sigma;}}_y}=\frac{1}{\sqrt{3}}(1+\frac{k^2}{{(r/r_i)}^2})[\frac{k_j^2}{k^2}-(1-\frac{k_j^2}{k^2}+\ln k_j^2)\frac{1}{k^2-1}]=(1+\frac{k^2}{{(r/r_i)}^2})\frac{{\mathrm{\&sigma;}}_z\&prime;}{{\mathrm{\&sigma;}}_y}---\left(6a\right)$

Similarly, by formula (1a)-(3a) or (4a)-(6a) can get elastic-plastic (r=r at the interface _j) residual stress, press p at r in adding _jThe stress that the place causes is used fourth strength theory and can be got r _jThe equivalent stress of place's total stress:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ej}}}{{\mathrm{\&sigma;}}_y}=1-\frac{1}{k^2-1}(\frac{k^2}{k_j^2}-1+\frac{2k^2}{k_j^2}\ln k_j)+\frac{\sqrt{3}{k}^2}{(k^2-1)k_j^2}\frac{p}{{\mathrm{\&sigma;}}_y}---\left(14b\right)$

Can get the container inner wall face (r=r of place by formula (1)-(3) _i) residual stress:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}\&prime;}{{\mathrm{\&sigma;}}_y}=-\frac{1}{\sqrt{3}(k^2-1)}(1-k_j^2+2k^2\ln k_j)---\left(15a\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ti}}\&prime;}{{\mathrm{\&sigma;}}_y}=-\frac{2}{\sqrt{3}(k^2-1)}(1-k_j^2+2\ln k_j)=2\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}\&prime;}{{\mathrm{\&sigma;}}_y}---\left(16a\right)$

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ri}}\&prime;}{{\mathrm{\&sigma;}}_y}=0---\left(17a\right)$

Press fourth strength theory, r _iThe equivalent stress of place's residual stress is:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;}{{\mathrm{\&sigma;}}_y}=-\frac{1}{k^2-1}(1-k_j^2+{2k}^2\ln k_j)---\left(18a\right)$

Just the same with result by the 3rd intensity theory, so k _jInevitable the same with the relation of k, promptly must be formula (23), so, determine k no matter press the 3rd or fourth strength theory _jFormula all be the same.

Can get r with same step _iThe equivalent stress of place's total stress is:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}}{{\mathrm{\&sigma;}}_y}=\sqrt{3}\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}\&prime;}{{\mathrm{\&sigma;}}_y}+\sqrt{3}{k}^2\frac{{\mathrm{\&sigma;}}_{\mathrm{zi}}^p}{{\mathrm{\&sigma;}}_y}=\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;}{{\mathrm{\&sigma;}}_y}+\frac{\sqrt{3}{k}^2}{k^2-1}\frac{p}{{\mathrm{\&sigma;}}_y}---\left(19a\right)$

Equally, (18a) can make p/ σ by formula (19a) _y=0 obtains.

Passing through type (18a), formula (14b) can be write as:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ej}}}{{\mathrm{\&sigma;}}_y}=\frac{{k_j}^2-1}{{k_j}^2}+\frac{\sqrt{3}{k}^2}{{k_j}^2(k^2-1)}\frac{p}{{\mathrm{\&sigma;}}_y}+\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;/{\mathrm{\&sigma;}}_y}{{k_j}^2}---\left(14c\right)$

Equally, make formula (1a)=formula (2a) and formula (1a)=formula (3a) all get:

${\left(\frac{r}{r_i}\right)}^2=k_r^2=\frac{1}{k^2-1}(k^2-k_j^2+k^2\ln {k_j}^2)=\left|\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;}{{\mathrm{\&sigma;}}_y}\right|+1$

Also just the same with result by the 3rd intensity theory.

Make p/ σ in formula (14b) or (14c) _y=0 gets r _jThe equivalent stress of the residual stress at place:

$\frac{{\mathrm{\&sigma;}}_{\mathrm{ej}}\&prime;}{{\mathrm{\&sigma;}}_y}=1-\frac{1}{k^2-1}(\frac{k^2}{k_j^2}-1+\frac{2k^2}{k_j^2}\ln k_j)=\frac{{k_j}^2-1}{{k_j}^2}+\frac{{\mathrm{\&sigma;}}_{\mathrm{ei}}\&prime;/{\mathrm{\&sigma;}}_y}{{k_j}^2}--\left(20a\right)$

Just the same with result again by the 3rd intensity theory.

In formula (14b), make σ _Ej/ σ _y=1 must the plastic zone degree of depth be k _jPressurized container work as r _jThe load that the place can bear when just having begun to surrender:

$\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{1}{\sqrt{3}{k}^2}(k^2-{k_j}^2+2k^2\ln k_j)---\left(24a\right)$

Formula (23) is write as

2k ²Lnk _j-k _j ²=k ²-2 substitution formulas (24a):

$\frac{p}{{\mathrm{\&sigma;}}_y}=2\frac{k^2-1}{\sqrt{3}{k}^2}=2\frac{p_e}{{\mathrm{\&sigma;}}_y}$ k≥2.2184574899167 (25a)

The bearing capacity formula that the self-reinforcing pressure vessel of safety that Here it is is pressed fourth strength theory, its image as shown in Figure 4.Do some explanations below convolution (25a) and the image thereof:

(1) curve ona (k=1～2.2184574899167).Consult the curve ofa of the 3rd intensity.

(2) curve aq (k 〉=2.2184574899167).Consult the curve ab of the 3rd intensity.This stage: σ _Ei'/σ _y=-1 (is equivalent to k ²Lnk _j ²-k ²-k _j ²+ 2=0), σ _Ej=σ _yσ _Ejmin, while σ _Ei=σ _yBearing capacity is: $\frac{p}{{\mathrm{\&sigma;}}_y}=2\frac{k^2-1}{\sqrt{3}{k}^2}=2\frac{p_e}{{\mathrm{\&sigma;}}_y}$

P/ σ _yIncrease along with the k increase (and k _jReduce along with the k increase); During k → ∞, k _j→ exp (0.5),

This stage bearing capacity greater than the os section less than the av section.

(3) dotted line av (k 〉=2.2184574899167): $\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{p_y}{{\mathrm{\&sigma;}}_y}=\frac{2}{\sqrt{3}}\ln k.$

This section curve and solid line ona are same, but behind k 〉=2.2184574899167, p/ σ _yAnd can't help av and determine.

(4) dotted line os (k=1～∞).Non-self intensification container maximum flexibility bearing capacity $\frac{p_e}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{\sqrt{3}{k}^2},$ It is a correlation curve.

(5) dotted line owa (k=1～2.2184574899167).With solid line aq be same $(\frac{p}{{\mathrm{\&sigma;}}_y}=2\frac{k^2-1}{\sqrt{3}{k}^2}),$ Bearing capacity on this section compares p _y/ σ _yAlso big.But in k=1～2.2184574899167 stages, bearing capacity be can't help curve owa and by curve ona, determined, otherwise σ _EiWith σ _EjAll will be above σ _y, dangerous.

K＜2.2184574899167 o'clock, | σ _Ei'/σ _y| permanent in 1.

Therefore, curve onaq is best bearing capacity curve.

Know that from above analysis and Fig. 3,4 a point is the intersection point of full yield line and best bearing capacity curve, have at this point:

Press the 3rd intensity theory: $\frac{p}{{\mathrm{\&sigma;}}_y}=\frac{k^2-1}{k^2}=\ln k$

Press fourth strength theory: $\frac{p}{{\mathrm{\&sigma;}}_y}=2\frac{k^2-1}{\sqrt{3}{k}^2}=\frac{2}{\sqrt{3}}\ln k$

Clearly, more than two formulas be of equal value, and the arrangement the result all be formula (22).

The theoretical foundation and the foundation of reference when certain law, relation and the data that obtain in the above analytic demonstration process, chart etc. can be used as the pressurized container engineering design also make theoretical each relationship between parameters of self intensification and Changing Pattern more clear, thorough and practical.

Claims (2)

1. the self-reinforcing pressure vessel of a safety is characterized in that: the physical dimension of this kind pressurized container and bearing capacity determine that by particular requirement specifically: its wall thickness by the 3rd intensity theory is

By fourth strength theory be

It is directly than by the 3rd intensity theory being

By fourth strength theory be

For k greater than by formula

The self-reinforcing pressure vessel of the value of determining, the plastic zone degree of depth of its wall thickness is pressed

Determine that its bearing capacity by the 3rd intensity theory is By fourth strength theory be

R wherein _iBe the internal diameter of self-reinforcing pressure vessel, k is the self-reinforcing pressure vessel outer radius and the ratio of inside radius, k _jBe self-reinforcing pressure vessel elasticity and the ratio of plastic zone interface radius with inside radius, σ _yBe the self-reinforcing pressure vessel YIELD STRENGTH, p is the interior pressure that self-reinforcing pressure vessel bore, p _eMaximum flexibility bearing capacity for non-self-reinforcing pressure vessel.

2. safe self-reinforcing pressure vessel as claimed in claim 1 is characterized in that: for k less than by formula

The self-reinforcing pressure vessel of the value of determining, its plastic zone size scope can be whole wall thickness, and this moment, its maximum load-carrying capacity was pressed full yield pressure, promptly by the 3rd intensity theory was

By fourth strength theory be

It is directly than by the 3rd intensity theory being By fourth strength theory be Its wall thickness by the 3rd intensity theory is

By fourth strength theory be

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