CN1723329A - Bore hole tool assembly, bearing system for use in such a bore hole tool assembly, and method of designing such a bore hole tool assembly - Google Patents

Abstract

Bore hole tool assembly comprising a housing and a mandrel reaching along an axial direction into the housing and a bearing system for transferring an axial load between the housing and the mandrel and allowing rotation of the housing relative to and about the mandrel, wherein the bearing system comprises at least two bearing stages each arranged to transfer part of the axial load, whereby each bearing stage comprises bearing means and mitigation means for distributing the load substantially proportionally over the respective bearing stages.

Description

The bearing arrangement that bore hole tool assembly, this class bore hole tool assembly of confession are used and the method for designing of this class bore hole tool assembly

Technical field

The present invention relates to a kind of bore hole tool assembly, particularly supply the boring bore hole tool assembly of hydrocarbon well, comprise axle and that a sleeve pipe and stretches into sleeve pipe vertically be used between sleeve pipe and axle, transmitting axial load and make sleeve pipe can be relatively, center on the bearing arrangement that axle is rotated, wherein bearing arrangement comprises at least two bearing levels, each is disposed so that the transmitting portions axial load, and each bearing level all comprises bearing arrangement.

In addition, the present invention relates to bearing arrangement and a kind of method that designs this class bore hole tool assembly that this class bore hole tool assembly of a kind of confession is used.

Background technology

In well bore and well completion operations, adopt the bore hole tool assembly that is shaped as tubular strings.These drill strings can comprise tube-like piece such as drilling rod, stem, drill collar, bushing pipe, sleeve pipe and the pipe of all kinds, size.These parts can be adopted with various configurations.

These drill strings bear mechanical tension and compression because of drill string weight with the friction of well bore wall.Rotate if need make one one of drill string be independent of the drill string remainder, just need a bearing assembly.This class bearing assembly bear with drill string in around the identical Tensile or Compressive Loading of pipe fitting.

In this class was used, bearing assembly normally was arranged in the utilizable annulus, this space be in the axle effect a drill string part pipe end and play between the pipe end of another drill string part of sleeve pipe effect.

Axle section and sleeve pipe section often require to be restricted because of practicality.For example, be subjected to the restriction of following needs for well with the maximum section of pipe, but promptly between well and pipe, need have an annulus to make the drilling fluids Channel Group that contains borehole cuttings.Minimum cross section may be subjected to the restriction of following needs, but promptly need make drilling fluids Channel Group or make intended size object such as gluing plug (cementationplugs) or wire rope configuration tool to pass through.In addition, for enough intensity is provided, the minimum wall thickness (MINI W.) of center roller and sleeve pipe also has requirement.

These requirements have limited the free space that bearing assembly is installed in annulus.Be contained in the single bearing level possibility insufficient strength in the free space, can't bear the load that bore hole tool assembly bears.Therefore, advised adopting have a plurality of bearing levels to disposed so that the bearing arrangement of transmitting portions axial load parallel to each other.

In fact, each bearing level is configured in respectively in the annulus, and an axial displacement is arranged relative to each other, and they are being worked axial load when sleeve pipe is delivered to axle in parallel with each other.

Have found that, can be in each bearing level of bearing arrangement in the place of the certain maximum static of inefficacy front transfer, bearing series just lost efficacy under a static load, and this static load significantly is lower than each bearing level peak load of mentioning and multiply by bearing progression in the series.Moreover under the rotation condition, the working life of system significantly is lower than the value by the working life that the indicates expection of single bearing.

Summary of the invention

Like this, one object of the present invention is exactly to improve the particularly bearable load of bearing arrangement of bore hole tool assembly.

Another object of the present invention provides a kind of bore hole tool assembly and this class bore hole tool assembly bearing arrangement of a kind of confession, and they have longer working life.

One or more purposes of the present invention realize by a kind of bore hole tool assembly, this assembly comprise axle and that a sleeve pipe and stretches into sleeve pipe vertically be used between sleeve pipe and axle, transmitting axial load and make sleeve pipe can be relatively, center on the bearing arrangement that axle is rotated, wherein bearing arrangement comprises at least two bearing levels, each is disposed so that the transmitting portions axial load, each bearing level comprises bearing arrangement and reliever, is used for load is distributed on each bearing level basically pro rata.

The present invention is based upon on the following understanding basis, promptly in fact axle and sleeve pipe each all be equivalent to have the member of the elastically deformable of certain mechanical stiffness.In order to describe the present invention, the entity mechanical stiffness is defined in the power of transmitting by a member on the certain-length and the increment of this length or the ratio of decrement.Be load skewness on bearings at different levels due to the elastic behavior of keeping away axle, sleeve pipe and bearing, adopt reliever to make that load is proportional basically to be distributed on each bearing level.Basically avoided some bearing level carrying to surpass its bearable load thus, and other bearing levels can also be born more load in the series.

Concerning present patent application, full payload proportional substantially distribution on bearings at different levels is understood as that and comprises a kind of distribution, the load that promptly acts on the bearing level of bearing the full payload largest portion is no more than 1.5 times of average loads, and average load is that full payload is divided by bearing arrangement middle (center) bearing progression.Preferably it is no more than 1.3 times of average loads, and more preferably it is no more than 1.1 times of average loads.

For fear of on a bearing level, surpassing peak load too early, below be sufficient, when promptly having only load on acting on one or more bearing levels to reach bearable load, just dispose reliever load is distributed on the corresponding bearing level substantially pro rata.

As long as since there is a bearing level to lose efficacy whole bearing arrangement with regard to inefficacy, working life also obtains prolonging from the heaviest bearing level alleviation of load because of load simultaneously.

In one embodiment, reliever comprises being disposed so that connect with corresponding bearing arrangement and transmits the deformable device of portion's axial load at least.Deformable device is out of shape under a load, and the load in the bearing level of super so reasonable carrying just part alleviates, and has turned to the distortion of anamorphic attachment for cinemascope, and the result just forces residue bearing level to bear more most relatively load.

Deformable device can comprise the fluid bellow net of the flexibility of fluidic intercommunication, and each bearing level comprises such flexible fluid bellows.Or alternatively, but deformable device can be made by the combined material of elasticity and plastic strain or by being essentially deformable material.

But the benefit with reliever of elastically deformable or elasticity and plastic strain is and can makes the reliever elastic biasing against bearing arrangement, thereby bearing arrangement is remained valid in the load range of top load in being in a ratio of with bearable load comprising.Top load is proportional substantially in the middle of same when being distributed on each bearing level, and the distribution of the wearing and tearing on the bearing arrangement on corresponding bearing level is just more even, and the overall life of whole like this bearing arrangement just further improves.

More preferably, the mechanical stiffness of anamorphic attachment for cinemascope is lower than the rigidity of sleeve pipe and by the rigidity of the axle of this grade overlap joint in one-level bearing at least.But the incompatible adequate compensation thus between sleeve pipe and axle elastic performance.

It is desirable to, if the mechanical stiffness of each bearing level is lower than the mechanical stiffness of sleeve pipe and is lower than the mechanical stiffness of axle, each bearing level is low about 2 to 2.5 times at least in the system, and then axial load is easy to proportional being distributed on the available bearing level most.Yet the axial length that whole like this bearing arrangement requires is longer, usually this to be difficult for accomplishing also be undesirable.

Therefore preferable is that the mechanical stiffness of deformable device is different from the rigidity of deformable device in other bearing levels in size in a bearing level.Axial load by each bearing level transmission just can be distributed on each bearing level again flexibly thus, because be lower than the axial load of the bearing level carrying with the mechanical stiffness that equates with other bearing levels by the load to axial of the lower bearing level carrying of mechanical stiffness.Like this, by otherwise will bear in the heaviest bearing level minimum mechanical stiffness is provided, the distribution of load on each bearing level is just more balanced.

Just reduced the required axial length of bearing arrangement thus.In fact, in each bearing level mechanical stiffness can be higher than sleeve pipe and/or axle rigidity 1/10th.

According to a second aspect of the present invention, provide a kind of for the bore hole tool assembly bearing arrangement used of oil field instrument assembly particularly, but be used between sleeve pipe and the axle that stretches into sleeve pipe vertically, transmitting axial load and make sleeve pipe center roller and rotate mutually around axle, this bearing arrangement comprises at least two bearing levels, each is disposed so that the transmitting portions axial load, each bearing level comprises bearing arrangement and reliever, is used for load is distributed on the corresponding bearing level basically pro rata.

Advantage according to this bearing arrangement of the present invention is introduced in the description of relevant above-mentioned hole instrument assembly with other embodiment.

According to a third aspect of the present invention, a kind of method according to the above design bore hole tool assembly is provided, this method may further comprise the steps:

A) selection is fit to the clutch shaft bearing level rigidity of a bearing level;

B) just extend in each bearing inter-stage sleeve pipe each several part given rigidity value and extend in the given rigidity value of each several part of the axle of each bearing inter-stage, calculate the value that is fit to the second bearing level rigidity, thereby axial load is distributed on the corresponding bearing level pro rata, selected clutch shaft bearing level rigidity value is used as input here.

Use the method, just can avoid experimental model is carried out trouble and convention of gathering through test consuming time.

This method can be the part of the method for producing bore hole tool assembly, wherein according to calculating, comprises that the bearing arrangement of each bearing level and axle, sleeve pipe are selected, assembled.

Particularly the two has along the occasion of the constant mechanical stiffness of each bearing level at axle and sleeve pipe, be fit to the required value of the mechanical stiffness of residue bearing level by expediently with clutch shaft bearing level rigidity with overlap ratio of rigidity between the pipe stiffness and the rigidity between clutch shaft bearing level rigidity and the axle rigidity and recently expressed.

Description of drawings

Be described in further detail the present invention below with reference to the accompanying drawings by way of example, in the accompanying drawings:

Fig. 1 schematically illustrates the sectional elevation of a bore hole tool assembly;

Fig. 2 is in stretchings (a portion) and a kind of elastic construction model that compresses under (b portion) state for bore hole tool assembly shown in Figure 1.

Similar reference symbol is at similar component among the figure.

The specific embodiment

With reference to figure 1, schematically show bore hole tool assembly 1 with sectional elevation.Bore hole tool assembly comprises a sleeve pipe 2 and a tubular mandrel 3, and axle 3 axially stretches in the sleeve pipe 2, stays an annulus between axle 3 and sleeve pipe 2.Several bearing levels (8,9) are set in the annulus, and thus, each bearing level one end is connected to the axle other end and is connected to sleeve pipe.Bearing level 8 is disposed so that transmit the axial tension of 2 of axle 3 and sleeve pipes in parallel with each other; Bearing level 9 is disposed so that transmit the compressive force of 3 of sleeve pipe 2 and axles.

Still with reference to figure 1, bearing level (8,9) comprises bearing arrangement (4,6) and reliever (5,7) for every grade.Kinds of bearings is not limit, and can be roller bearing, ball bearing, spherical roller bearing, conical roller bearing, frustum of a cone roller bearing.The elongated bearing of preferentially selecting for use the relative sleeve pipe 2 and the rotating shaft of axle 3 radially to dispose is because the load in this bearing is to distribute on the major part of bearing surface.Reliever can any deformable material such as the form of plastics or rubber one class material or be provided with the form of stage clip such as annular disc spring.As shown, reliever is provided so that to connect with the corresponding axis bearing apparatus in this grade and transmits load in each bearing level.

By connecting with bearing arrangement deformable material is set, the mechanical stiffness of the relative bearing arrangement of the mechanical stiffness of bearing level itself decreases, and makes the distribution on available bearing level controlled by lower mechanical stiffness is provided to a bearing level.To not bear the load higher if there is this bearing level of reliever than reasonable load.This mechanical stiffness is lower than the rigidity of such bearing level, if promptly there is not reliever, this bearing level will be born the load lower than reasonable load.

Bore hole tool assembly shown in Figure 1 has the bearing arrangement that contains four bearing levels.Have found that when being used for typical hydrocarbon well, four bearing levels can provide the collocation of the best between assembly intensity, reliability, life-span, the size.But be not difficult to find out that it is any number of 2 that bearing progression can change over minimum value.

Further illustrate the method that designs bore hole tool assembly by the present invention below with reference to a kind of mathematical method.Mathematical method is applicable to that generally quantity arbitrarily and the bearing level of representing with n.What mention is that the result of the mathematical method of discussing below all is suitable for drawing high with compressive load, also is applicable to the occasion that makes continuous deformable device face toward pretension each other.

Fig. 2 illustrates an elastic construction model of bore hole tool assembly shown in Figure 1, and the corresponding boring bar tool of Fig. 2 a bears to draw and carries, and the corresponding drilling tool of Fig. 2 b bears ballast.Each bearing level is numbered with ordinal number k, k=1, and 2 ..., n.Ordinal number k=1 refers to that this locates whole load (draw or press) and act on clutch shaft bearing level in the axle, and index k=n refers to that this locates whole load (draw or press) and act on last bearing level in the sleeve pipe.

Sleeve pipe and axle can be made of the different sections with different length for example or different section area.Each this class section is represented with ordinal number i.Section adds up to 2n.Each this class section has a mechanical stiffness, and this rigidity is defined as power and the increment of this length or the ratio of decrement that the section through having certain-length transmits.Rigidity value K along the sleeve pipe respective section _HiExpression, i=1,2 ..., 2n.Rigidity value K along the axle respective section _MiExpression, i=1,2 ..., 2n.

Along the displacement of gauge point between axle and sleeve pipe section by U _jProvide, j=1,2,3 ..., (4n+2).Even number j is corresponding to the point on the sleeve pipe, and odd number j is corresponding to the point on the axle.

The axle stiffness K of certain section (i number) _MiWith the sleeve pipe stiffness K _HiAvailable following equation is calculated based on cross dimensions and material property:

$K_{\mathrm{hi}}=\frac{A_{\mathrm{hi}}{E}_h}{L_i},$

$K_{\mathrm{mi}}=\frac{A_{\mathrm{mi}}{E}_m}{L_i}$

Wherein

A _HiCross-sectional area [m for sleeve pipe section i ²];

A _MiCross-sectional area [m for axle section i ²];

L _iLength [m] for sleeve pipe or axle section i;

E _hModulus of elasticity [N/m for shell material ²];

E _mModulus of elasticity [N/m for the axle material ²].

Bearing level stiffness K ₁, K ₂..., K _nCorresponding to the composite rigidity that bearing and deformable device are together in series in bearing level k, deformable device system represents with the form of spring in this example.

K wherein _{Bearing, k}And K _{Spring, k}Be respectively the bearing rigidity and the spring rate of k bearing level.

A task will find each spring rate K exactly _{Spring, k}Value so that the total axial force F that transmits between axle and sleeve pipe is distributed on the available bearing level by ratio.In other words, the power F of each bearing level transmission _kUnder the ideal state must be for the 1/n of total power $F_k=\frac{1}{n}F.$

As the proportional distribution of power, along the displacement U of axle _j(see figure 2) just becomes:

U ₁=0。(as benchmark)

${\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3=\frac{\mathrm{<figure-callout\; id="1"\; label="F\; K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_m},$

${\mathrm{<figure-callout\; id="5"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_5={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{n-1}{n}\frac{\mathrm{<figure-callout\; id="2"\; label="F\; K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_m}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}\left[(n-1)\frac{1}{K_{m2}}\right]F,$

${\mathrm{<figure-callout\; id="7"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_7={\mathrm{<figure-callout\; id="5"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_5+\frac{\mathrm{<figure-callout\; id="3"\; label="F\; K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_m}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}\left[(n-1)(\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m3}})\right]F,$

${\mathrm{<figure-callout\; id="9"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_9=U_7(1-\frac{2}{n})\frac{\mathrm{<figure-callout\; id="4"\; label="F\; K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_m}{=\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[(n-1)(\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m3}})+(n-2)\left(\frac{1}{K_{m4}}\right)]F,$

$U_{11}={\mathrm{<figure-callout\; id="9"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_9+(1-\frac{2}{n})\frac{\mathrm{<figure-callout\; id="5"\; label="F\; K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_m}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[(n-1)(\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m3}})+(n-2)(\frac{1}{{\mathrm{<figure-callout\; id="4"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m5}})]F,$

$U_{13}=U_{11}(1-\frac{3}{n})\frac{\mathrm{<figure-callout\; id="6"\; label="F\; K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_m}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[(n-1)(\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m3}})+(n-2)(\frac{1}{{\mathrm{<figure-callout\; id="4"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m5}})+(n-3)\left(\frac{1}{K_{m6}}\right)]F,$

$U_{15}=U_{13}+(1-\frac{3}{n})\frac{\mathrm{<figure-callout\; id="7"\; label="F\; K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_m}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[(n-1)(\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m3}})+(n-2)(\frac{1}{{\mathrm{<figure-callout\; id="4"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m5}})+(n-3)(\frac{1}{{\mathrm{<figure-callout\; id="6"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{1}{K_{m7}})]F,$

Or the like.

Select clutch shaft bearing level rigidity value K ₁Usually, the rigidity of bearing arrangement has a set-point, and the reliever that has a suitable rigidity value by suitable selection obtains the rigidity of selected bearing level.Can consider to select suitable bearing level rigidity based on intensity.Under the expectation situation of the proportional distribution of power, along the displacement components u of sleeve pipe _iAvailable K ₁Expression:

${\mathrm{<figure-callout\; id="2"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_2={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{K_1}\frac{F}{n},$

${\mathrm{<figure-callout\; id="4"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_4={\mathrm{<figure-callout\; id="2"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_2+\frac{1}{n}\frac{\mathrm{<figure-callout\; id="1"\; label="F\; K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_h}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}+\frac{1}{K_{h1}}]F,$

${\mathrm{<figure-callout\; id="6"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_6={\mathrm{<figure-callout\; id="4"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_4+\frac{1}{n}\frac{\mathrm{<figure-callout\; id="2"\; label="F\; K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_h}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}+\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{1}{K_{h2}}]F,$

${\mathrm{<figure-callout\; id="8"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_8={\mathrm{<figure-callout\; id="6"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_6+\frac{2}{n}\frac{\mathrm{<figure-callout\; id="2"\; label="F\; K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_h}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}+\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{2}{K_{h3}}]F,$

$U_{10}=U_a+\frac{2}{n}\frac{\mathrm{<figure-callout\; id="4"\; label="F\; K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_h}{=\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}+\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{2}{{\mathrm{<figure-callout\; id="3"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{2}{K_{h4}}]F,$

$U_{12}=U_{10}+\frac{3}{n}\frac{\mathrm{<figure-callout\; id="5"\; label="F\; K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_h}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}+\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{2}{{\mathrm{<figure-callout\; id="3"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{2}{{\mathrm{<figure-callout\; id="4"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{3}{K_{h5}}]F,$

$U_{14}=U_{12}+\frac{3}{n}\frac{\mathrm{<figure-callout\; id="5"\; label="F\; K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}{K_h}={\mathrm{<figure-callout\; id="3"\; label="U"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">U</figure-callout>}}_3+\frac{1}{n}[\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}+\frac{1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{2}{{\mathrm{<figure-callout\; id="3"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{2}{{\mathrm{<figure-callout\; id="4"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{3}{{\mathrm{<figure-callout\; id="5"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{3}{K_{h6}}]F,$

Or the like.

Chosen rigidity value K ₁After, the optimum value K of suitable all the other bearing level rigidity ₂, K ₃..., K _nSolve with regard to available mathematical way, evenly distribute to exert all one's strength.Its method is as follows.For making the second bearing level rigidity become the best, should keep:

${\mathrm{<figure-callout\; id="2"\; label="F"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}_2=K_2(U_8-U_7)\&equiv;\frac{1}{n}F,$

${\mathrm{<figure-callout\; id="3"\; label="F"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}_3=K_3(U_{10}-U_{11})\&equiv;\frac{1}{n}F$

${\mathrm{<figure-callout\; id="4"\; label="F"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">F</figure-callout>}}_4=K_4(U_{14}-U_{15})\&equiv;\frac{1}{n}F,$

Or the like

With the displacement U of the expression formula replacement that provides above along axle (odd number j) and sleeve pipe (even number j) _jThe time, the best ratio of rigidity K of bearing level _k/ K ₁, k=2,3 ..., n, this just obtains:

$\frac{K_2}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+(\frac{K_1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{K_1}{K_{h2}})-(n-1)(\frac{K_1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{K_1}{K_{m3}})]}^{-1},$

$\frac{K_3}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+(\frac{K_1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{K_1}{K_{h2}})+2(\frac{K_1}{{\mathrm{<figure-callout\; id="3"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{K_1}{K_{h4}})-(n-1)(\frac{K_1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{K_1}{K_{m3}})-(n-2)(\frac{K_1}{{\mathrm{<figure-callout\; id="4"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{K_1}{K_{m5}})]}^{-1},$

$\frac{K_4}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={\left[\begin{array}{c}1+(\frac{K_1}{{\mathrm{<figure-callout\; id="1"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{K_1}{K_{h2}})+2(\frac{K_1}{{\mathrm{<figure-callout\; id="3"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{K_1}{K_{h4}})+3(\frac{K_1}{{\mathrm{<figure-callout\; id="5"\; label="K\; h"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_h}+\frac{K_1}{K_{h8}})...\\ ...-(n-1)(\frac{K_1}{{\mathrm{<figure-callout\; id="2"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{K_1}{K_{m3}})-(n-2)(\frac{K_1}{{\mathrm{<figure-callout\; id="4"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{K_1}{K_{m5}})-(n-3)(\frac{K_1}{{\mathrm{<figure-callout\; id="6"\; label="K\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_m}+\frac{K_1}{K_{m7}})\end{array}\right]}^{-1},$

Or the like.

These equations are applicable to the pervasive situation that rigidity value can be different in each section i sleeve pipe and the axle.But to I'm afraid more real situation, promptly sleeve pipe and axle rigidity are constant along each section, also promptly establish center roller K _Ml=K _M2=...=K _Mn≡ K _m, for sleeve pipe K _Hl=K _H2=...=K _Hn≡ K _h, equation just can be simplified greatly.Best bearing level rigidity like this is just provided by following formula:

$\frac{{\mathrm{<figure-callout\; id="1"\; label="K\; k\; K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_k}{K_{}}={[1+A_k\frac{K_1}{K_h}+B_k\frac{K_1}{K_m}]}^{-1},$

A _k＝k(k-1)

B _k＝(k-1)(k-2n)

As seen, the proportional best ratio of rigidity K that is distributed on the available bearing level that exerts all one's strength _k/ K ₁Only depend on ratio K ₁/ K _hWith K ₁/ K _mWhen clutch shaft bearing level rigidity was selected lowly, the difference of these best rigidity just reduced.

Should be mentioned that entity stiffness comprises K _h, K _m, K ₁, K _k, be to provide with the dimension of per unit apart from power [N/m].Factor A _kAnd B _kAnd ordinal number then is nondimensional.

With regard to each bearing level in the bearing arrangement, if the mechanical stiffness of deformable reliever is lower than 2 to 2.5 times of the mechanical stiffness of sleeve pipe or the mechanical stiffnesses of axle (no matter which is a reckling among both) in each bearing level, axial load just is easy to most proportional being distributed on the available bearing level so, because the ratio between different bearing level rigidity reaches unanimity.

But owing to be subjected to the restriction of size and requirement of strength one class in the practical application, in each bearing level the mechanical stiffness of deformable reliever be in than sleeve pipe or the low person of axle rigidity value, each bearing level low 0.25 to 2.5 times somewhere.So, exerting all one's strength is distributed on the available bearing level more pro rata, can obtain by making corresponding bearing level rigidity optimization.More preferably, do not cause a single bearing level to bear load thus, for instance 1.5 * F/n at the most significantly greater than the certain proportion of full payload.

Bearing arrangement is provided with two bearing levels and constant with the cover pipe stiffness along axle, then concerning bore hole tool assembly with this bearing arrangement, for exert all one's strength proportional be distributed in choose on two bearing levels and K ₁The best stiffness K of coupling ₂Just provide by following formula:

$\frac{K_2}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+2\frac{K_1}{K_h}-2\frac{K_1}{K_m}]}^{-1}$

Be provided with three bearing levels and along main shaft and cover pipe stiffness constant bore hole tool assembly, best stiffness K for its bearing arrangement ₂And K ₃For:

$\frac{K_2}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+2\frac{K_1}{K_h}-4\frac{K_1}{K_m}]}^{-1},$

$\frac{K_3}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+6\frac{K_1}{K_h}-6\frac{K_1}{K_m}]}^{-1}$

Be provided with four bearing levels and along axle and cover pipe stiffness constant bore hole tool assembly, best stiffness K as for its bearing arrangement ₂, K ₃And K ₄For:

$\frac{K_2}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+2\frac{K_1}{K_h}-6\frac{K_1}{K_m}]}^{-1},$

$\frac{K_3}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+6\frac{K_1}{K_h}-10\frac{K_1}{K_m}]}^{-1},$

$\frac{K_4}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+12\frac{K_1}{K_h}-12\frac{K_1}{K_m}]}^{-1}$

Imagined already in the above description, it is linear that the relation of mechanical stiffness and load is.If member is non-linear to the mechanical response of a load, just should determine the mechanical stiffness of this mechanism according to the load that is occurred, whether the abundant proportional distribution on bearings at different levels of bearing arrangement general design load.

This class physical constraints of availability that it should be understood that spring can cause a kind of design, some section length L of its middle sleeve/axle _iNeed length, so that obtain desirable bearing level rigidity than other.The equation that provides so, above must adapt to this particular case.

Example

Imagine an oil well drill pipe and can transmit the 2000kN Tensile or Compressive Loading.Need total power is evenly distributed on n=4 the bearing level.So just do not surpass the static load intensity of any bearing too early, the working life of bearing arrangement also improves.Drilling tool is based on a cylindrical mandrel and a circular casing.

Particularly cross-sectional area is restricted for the functional requirement center roller of design and the size of sleeve pipe.Axle and casing steel are made, and modulus of elasticity is E _m=E _h=2.1 * 10 ⁵N/mm ²The cover external diameter of pipe is restricted to maximum 181mm.The cylindrical mandrel internal diameter is restricted to minimum 63.5mm.

Adopted thrust bearing, (axle) footpath is 110mm in it, and outer (cover) footpath is 145mm, and axial length is 25mm.Manufacturer is known by bearing, and the rigidity of all bearings is K _Bearing=4.11 * 10 ⁶N/mm, it is defined as being added to the ratio of the axial shortening amount of the bearing that axial force and the axial force that is added on the bearing on the bearing cause.Axle external diameter along each section of axle is 100mm, so just can make bearing adaptive and feel relieved in instrument.Casing inner diameter along each section of sleeve pipe is 150mm.Bearing, spring and be connected to axle and the needed section length (L of the fixture/joint of sleeve pipe along sleeve pipe and axle ₁, L ₂, L ₃Deng) be 75mm.

In order to make each bearing level energy transmitted load 500kN, be provided with a disc spring, its thickness is 20mm, rigidity is

K _Spring=1.35 * 10 ⁶N/mm.

Stiffness K along the every 75mm section of sleeve pipe _hProvide by following formula:

$K_h=\frac{E_h{A}_h}{L}.$

$=\frac{2.1\&times;{10}^5\&times;\frac{\mathrm{\&pi;}}{4}({181}^2-{150}^2)}{75}=22.6\&times;{10}^{6}N/\mathrm{<figure-callout\; id="2"\; label="m\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">m</figure-callout>}{m}^{}{}^2$

Stiffness K along the every 75mm section of axle _mProvide by following formula:

$K_m=\frac{E_m{A}_m}{L}$

$=\frac{2.1\&times;{10}^5\&times;\frac{\mathrm{\&pi;}}{4}({100}^2-{63.5}^2)}{75}=13.1\&times;{10}^{6}N/\mathrm{<figure-callout\; id="2"\; label="m\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">m</figure-callout>}{m}^{}{}^2$

At clutch shaft bearing level place, two disc springs are installed in series, and this more approaches rigidity is reduced by half.At the bearing of clutch shaft bearing level series connection and the composite rigidity K of two springs ₁Provide by following formula:

$=\frac{\frac{1}{2}\&times;1.35\&times;{10}^6\&times;4.11\&times;{10}^6}{\frac{1}{2}\&times;1.35\&times;{10}^6+4.11\&times;{10}^8}=0.58\&times;{10}^{6}N/\mathrm{<figure-callout\; id="2"\; label="m\; m"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">m</figure-callout>}{m}^{}{}^2$

Thereby make relevant ratio of rigidity K ₁/ K _hAnd K ₁/ K _mBecome:

$\frac{K_1}{K_h}=\frac{0.58\&times;{10}^6}{22.6\&times;{10}^6}=0.025,$

$\frac{K_1}{K_m}=\frac{0.58\&times;{10}^6}{13.1\&times;{10}^6}=0.044$

Utilize these values, can calculate the ratio of rigidity of other bearing levels, so that load is distributed on all bearing levels pro rata:

$\frac{K_2}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+2\frac{K_1}{K_h}-6\frac{K_1}{K_m}]}^{-1}={[1+2\&times;0.025-6\&times;0.044]}^{-1}=1.27\&DoubleRightArrow;{\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_2=1.27{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1$

$\frac{K_3}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+6\frac{K_1}{K_h}-10\frac{K_1}{K_m}]}^{-1}={[1+6\&times;0.025-10\&times;0.044]}^{-1}=1.16\&DoubleRightArrow;{\mathrm{<figure-callout\; id="3"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_3=1.16{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1$

$\frac{K_4}{{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1}={[1+12\frac{K_1}{K_n}-12\frac{K_1}{K_m}]}^{-1}={[1+12\&times;0.025-12\&times;0.044]}^{-1}=1.29\&DoubleRightArrow;{\mathrm{<figure-callout\; id="4"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_4=1.29{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A20038010568700161.png"\; state="\{\{state\}\}">K</figure-callout>}}_1$

By the harder spring of the ratio first order of connecting with each bearing is installed, then can obtain the value of higher bearing level rigidity.Be applicable to that bearing level k (series connection of two springs) locates composite rigidity K _kRelation can be expressed as follows again:

This moment, bearing level 2,3,4 required spring rate values were:

In this example, the difference of best rigidity value is smaller.This be because, by the design, the cross-sectional area of axle and sleeve pipe only differs from about 2 times, this is more desirable.In addition, get two springs and be series at and draw a bearing level, can make two important ratio K ₁/ K _mAnd K ₁/ K _hBe restricted.Select other designs for use, the difference of best spring rate value can enlarge markedly.

For some application, for example the above, the axle of setting is preferentially selected tubulose for use.However, adopt the sleeve pipe and/or the axle of solid shafting even employing noncircular cross section such as square section, the present invention also is effective.

Obviously, bearing arrangement described herein also can be used for the other technologies field valuably except that being used for bore hole tool assembly, desires to make sizable axial load to be delivered to axle or axle rotationally from sleeve pipe there, or back transfer.The example in these fields is included in particularly application, locomotive, the seagoing vessel driving shaft in the heavy truck of automobile.This also is the modification of described bore hole tool assembly method for designing.

Unpub No. 02080230 european patent application is included appendix in for the application's priority requisition and at this, and it has described an embodiment who is used for borehole systems in detail, and bore hole tool assembly of the present invention can obtain embodying therein valuably.

Claims (9)

1. A bore hole tool assembly it comprise the bearing arrangement that axle and that a sleeve pipe and stretches into this sleeve pipe vertically is used for transmitting the axial load between this sleeve pipe and this axle and this sleeve pipe can be rotated relative to this axle, around this axle, wherein this bearing arrangement comprises at least two bearing levels, each bearing level is configured to transmit the part of described axial load, and each bearing level all comprises and is used for load is distributed in reliever and bearing arrangement on the corresponding bearing level basically pro rata.
2. 2. bore hole tool assembly as claimed in claim 1 is characterized in that described reliever comprises the deformable device with a mechanical stiffness, and described deformable device is configured in series be passed to the described axial load of small part with the corresponding axis bearing apparatus.
3. 3. bore hole tool assembly as claimed in claim 2, wherein the mechanical stiffness of anamorphic attachment for cinemascope is lower than by the mechanical stiffness of the sleeve pipe in the section of this bearing level overlap joint and the mechanical stiffness of axle at least one bearing level.
4. 4. as claim 2 or 3 described bore hole tool assemblies, it is characterized in that the mechanical stiffness of deformable device is different from the mechanical stiffness of deformable device in other bearing levels or at least one other bearing level in size at least one bearing level.
5. 5. as the described bore hole tool assembly of any one aforementioned claim, it is characterized in that, differ from one another, differ less than 3 times at the mechanical stiffness of the mechanical stiffness of that part of sleeve pipe that extends between the bearing level and that part of axle of between the bearing level, extending.
6. 6. as the described bore hole tool assembly of any one aforementioned claim, it is characterized in that axle is a pipe end, preferably, sleeve pipe is a pipe end.
7. 7. one kind supplies bore hole tool assembly to use, but be used between sleeve pipe and the axle that stretches into sleeve pipe vertically, transmitting axial load and make sleeve pipe center roller, the bearing arrangement that rotates around axle mutually, this bearing arrangement comprises at least two bearing levels, each bearing level is configured to transmit the part of described axial load, and each bearing level comprises and is used for load is distributed in reliever and bearing arrangement on the corresponding bearing level basically pro rata.
8. 8. as the method for designing of any one described bore hole tool assembly in the claim 1 to 6, comprise the following steps:

   A) selection is used for the clutch shaft bearing level rigidity value of a bearing level;

   B) just at the given rigidity value of the given rigidity value of the sleeve pipe each several part that extends between the bearing level and the axle each several part that between the bearing level, extends, calculating is used for the value of the second bearing level rigidity, thus axial load is distributed on the corresponding bearing level basically pro rata, clutch shaft bearing level rigidity value is used as input here.
9. 9. method as claimed in claim 8, wherein step b comprises:

   B1) determine first ratio of rigidity, promptly described clutch shaft bearing level rigidity and the ratio that overlaps pipe stiffness;

   B2) determine second ratio of rigidity, the ratio of promptly described clutch shaft bearing level rigidity and axle rigidity.

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