CN104089770A - Asymmetric double-arm centrifugal loading device - Google Patents

Abstract

The invention relates to an asymmetric double-arm centrifugal loading device. The asymmetric double-arm centrifugal loading device comprises an upright and two side arms, wherein the upright is connected with a propeller hub, and the two side arms are installed on the upright. The tail end of each side arm is provided with a counter weight. When the two side arms and the counter weights on the two side arms rotate around a propeller hub rotating shaft through the upright, a generated twisting force bending moment and a generated thrust bending moment are opposite in direction, and a rotating blade moment and centrifugal force are the same in direction. The lengths of the two side arms and the included angles between the two side arms and the rotating shaft are adjusted so that the centrifugal force borne by the propeller hub, the thrust bending moment, the twisting force bending moment and the distance-adjusting rotating blade moment can be accurately simulated. In a centrifugal loading test process, the asymmetric double-arm centrifugal loading device can be used for accurately simulating the blade centrifugal force borne by the propeller hub, the thrust bending moment, the twisting force bending moment and the distance-adjusting rotating blade moment, wherein the thrust bending moment, the twisting force bending moment and the distance-adjusting rotating blade moment are in the three directions, and an average error is smaller than 1%. Thus, loading test verification can be effectively conducted on the strength of a propeller hub assembly, particularly a blade root bolt connected with the propeller hub and a blade.

Description

Asymmetric both arms centrifugal loading device

Technical field

The present invention relates to a kind of device of simulating the stress that screw propeller works in fluid, especially can accurately simulate the centrifugal loading device of the suffered centrifugal force of screw propeller and suffered all three yawning moments thereof.

Background technology

Propeller hub is the member that needs emphasis to consider in pitch propeller device design.The test that can effectively verify propeller hub Intensity Design mainly contains two kinds: the centrifugal loading pouring weight in full scale tests and land simulation test in water.In water, full scale tests acquired results is true, accurate and effective, but testing expenses are high, the test period long and require very high to water depth, flow velocity etc.Land centrifugal loading pouring weight simulation test, adopts centrifugal loading device to replace blade, as shown in Figure 1 and Figure 2, and the centrifugal force simulation suffered centrifugal force of blade working and the hydrodynamic force (square) of utilizing centrifugal loading device to produce in rotary course.Centrifugal load test is carried out on the testing table of land, and the difficulty of its test funds, test period and test is all much smaller than full scale tests in water.

At present, existing centrifugal loading device consists of weight, side arm and column three parts, is single armed, single heavy hammer structure.Centrifugal loading device is installed on propeller hub, and rotating shaft moves under working speed, and the centrifugal force that utilizes centrifugal loading device to produce can be simulated propeller hub actual suffered centrifugal force, thrust moment of flexure and torsion moment of flexure.Can the actual loading of propeller hub comprises centrifugal force, thrust, torsion and corresponding rotating vane moment of torsion, thrust moment of flexure and the torsion moment of flexure of blade, and accurately simulate in propeller hub actual loading in experimental test: the centrifugal force of blade and propeller hub the thrust moment of flexure, torsion moment of flexure and the rotating vane moment of torsion that are subject to, just determined can correctly effectively check, verify propeller hub intensity.Existing (single armed) centrifugal loading device can not be simulated thrust moment of flexure, torsion moment of flexure and rotating vane moment of torsion simultaneously, and therefore, along with the raising of testing requirements, existing centrifugal loading device can not meet testing requirements.

Summary of the invention

The present invention will provide a kind of asymmetric both arms centrifugal loading device, can overcome the deficiency that existing centrifugal loading device can not be simulated the suffered all moments of propeller hub; This asymmetric both arms centrifugal loading device can not only be simulated centrifugal force, thrust moment of flexure and torsion moment of flexure, and can also simulate rotating vane moment simultaneously, has the accuracy of higher controllability and the outer load of simulation.

For achieving the above object, technical scheme of the present invention is: a kind of asymmetric both arms centrifugal loading device, comprise a column being connected with propeller hub, two side arms that arrange on column, each side arm end respectively has a weight, be characterized in: two side arms and weight above thereof by column when propeller hub rotating shaft is rotated, the torsion moment of flexure and the thrust moment of flexure opposite direction that produce, rotating vane moment is identical with centrifugal force direction, adjust two side arms length and with the angle of rotating shaft, produce accurately suffered centrifugal force and the thrust moment of flexure of simulation propeller hub, torsion moment of flexure, three moments of roll adjustment rotating vane moment of torsion.

The length of two side arms and cross section are all not identical, and the axis of two side arms is not identical with propeller hub rotating shaft angulation.The length of two side arms and with the parameterisable adjustment respectively of the angle of propeller hub rotating shaft.

The invention has the beneficial effects as follows:

Two side arms are set on column of the present invention, and each side arm end respectively has a weight, for simulate propeller hub the asymmetry of stressed system, the length of two side arms and cross section are all not identical, and the axis of two side arms is not identical with rotating shaft angulation yet.When two side arms and weight above thereof rotate around the shaft, the torsion moment of flexure of generation is contrary with thrust moment of flexure symbol, and rotating vane moment is identical with centrifugal force symbol.Utilize couple principle, by adjust two side arms length and with the angle of rotating shaft, with regard to having solved traditional single armed charger, cannot simulate like this problem of rotating vane moment of torsion, thrust moment and torsion moment of flexure simultaneously.By the present invention, can produce the suffered centrifugal force of simulation propeller hub and thrust moment of flexure, torsion moment of flexure, three moments of roll adjustment rotating vane moment of torsion.

In centrifugal load test process, can utilize this charger accurately to simulate the moment of the suffered blade centrifugal force of (average error < 1%) propeller hub and thrust moment of flexure, torsion moment of flexure, all three directions of roll adjustment rotating vane moment of torsion.Thereby can, effectively to propeller hub assembly, particularly to connecting the intensity of the blade root bolt of propeller hub and blade, carry out load test checking.

Accompanying drawing explanation

Fig. 1 is the tuning for Controllable Pitch Propeller schematic diagram with blade;

Fig. 2 is the tuning for Controllable Pitch Propeller schematic diagram with asymmetric both arms centrifugal loading device;

Fig. 3 is the schematic diagram that asymmetric both arms centrifugal loading device of the present invention is connected with blade root flange;

Fig. 4 is fixed-axis rotation schematic diagram;

Fig. 5 is product of inertia translation transformation schematic diagram;

Fig. 6 is the rotation schematic diagram of coordinate system;

Fig. 7 is the structural front view of embodiments of the invention;

Fig. 8 is the vertical view of embodiments of the invention;

Fig. 9 is along A-A cut-open view in Fig. 8;

Figure 10 is along B-B cut-open view in Fig. 8.

Embodiment

Below in conjunction with accompanying drawing and embodiment, the invention will be further described.

As shown in Fig. 3,7,8,9,10, a kind of asymmetric both arms centrifugal loading device, comprises two side arms that arrange on a column 3 being connected with blade root flange 6, column 3.

The end of the first side arm 2 and the second side arm 4 has respectively the first weight 1 and the second weight 5, for simulate propeller hub the asymmetry of stressed system, the length of the first side arm 2 and the second side arm 4 and cross section are all not identical, and the axis of the first side arm 2 and the second side arm 4 is not identical with propeller hub rotating shaft 7 angulations yet.The length of two side arms and with the parameterisable adjustment respectively of the angle of propeller hub rotating shaft 7.During 7 rotation around the shaft of two side arms and weight above thereof, the torsion moment of flexure of generation and thrust moment of flexure symbol (direction) are contrary, and rotating vane moment is identical with centrifugal force symbol (direction).Utilize couple principle, by adjust two side arms length and with the angle of rotating shaft, with regard to having solved traditional single armed charger, cannot simulate like this problem of rotating vane moment of torsion, thrust moment of flexure and torsion moment of flexure that propeller hub is subject to simultaneously.By the present invention, can accurately simulate the actual centrifugal force being subject to of propeller hub and thrust moment of flexure, torsion moment of flexure, three moments of roll adjustment rotating vane moment of torsion.

Method for designing of the present invention:

The first step, sets up the parametrization computation model of single arm structure

One, the simplification of the inertial force system of fixed-axis rotation rigid body, as shown in Figure 4:

$\left\{\begin{array}{c}{F}_{\mathrm{Ix}}=m(x_c{\mathrm{\&omega;}}^2+y_c\mathrm{\&alpha;})\\ F_{\mathrm{Iy}}=m(y_c{\mathrm{\&omega;}}^2-x_c\mathrm{\&alpha;})\\ F_{\mathrm{Iz}}=0\\ M_{\mathrm{Ix}}=I_{\mathrm{xz}}\mathrm{\&alpha;}-I_{\mathrm{yz}}{\mathrm{\&omega;}}^2\\ M_{\mathrm{Iy}}=I_{\mathrm{yz}}\mathrm{\&alpha;}+I_{\mathrm{xz}}{\mathrm{\&omega;}}^2\\ M_{\mathrm{Iz}}=-I_z\mathrm{\&alpha;}\end{array}\right.---\left(1\right)$

In formula: the quality that m is rigid body; x _c, y _ccoordinate for rigid body barycenter; for the moment of inertia of rigid body to axle z; And

$\left\{\begin{array}{c}{I}_{\mathrm{xz}}=\mathrm{\&Sigma;}{m}_i{x}_i{z}_i\\ I_{\mathrm{yz}}=\mathrm{\&Sigma;}{m}_i{y}_i{z}_i\end{array}\right.$

Be called rigid body to axle x, z and axle y, the product of inertia of z.X in formula (1), y, z has rotation symmetry.

Two, the general formulae of product of inertia translation and rotational transform

1. the general type of product of inertia translation transformation

1) O ' be rigid body barycenter, i.e. x ' as shown in Figure 5, _c=0, y ' _c=0, z ' _c=0,

$\left\{\begin{array}{c}{I}_{\mathrm{xy}}={\mathrm{mx}}_c{y}_c+I_{x^{\&prime;}{y}^{\&prime;}}\\ I_{\mathrm{xz}}={\mathrm{mx}}_c{z}_c+I_{x^{\&prime;}{z}^{\&prime;}}\\ I_{\mathrm{yz}}={\mathrm{my}}_c{z}_c+I_{y^{\&prime;}{z}^{\&prime;}}\end{array}\right.---\left(2\right)$

And now O ' coordinate in Oxyz (x ' _o, y ' _o, z ' _o) coordinate (x of barycenter in Oxyz that be rigid body _c, y _c, z _c), therefore formula (3) can be expressed as again:

$\left\{\begin{array}{c}{I}_{\mathrm{xy}}={\mathrm{mx}}_{o^{\&prime;}}{y}_{o^{\&prime;}}+I_{x^{\&prime;}{y}^{\&prime;}}\\ I_{\mathrm{xz}}={\mathrm{mx}}_{o^{\&prime;}}{z}_{o^{\&prime;}}+I_{x^{\&prime;}{z}^{\&prime;}}\\ I_{\mathrm{yz}}={\mathrm{my}}_{o^{\&prime;}}{z}_{o^{\&prime;}}+I_{y^{\&prime;}{z}^{\&prime;}}\end{array}\right.---\left(3\right)$

2) principal axis of inertia that in O ' x ' y ' z ', each coordinate axis is rigid body, i.e. I _{x ' y '}=0, I _{x ' z '}=0, I _{y ' z '}=0:

$\left\{\begin{array}{c}{I}_{\mathrm{xy}}={\mathrm{mx}}_{o^{\&prime;}}{y}_{o^{\&prime;}}+{\mathrm{mx}}_{o^{\&prime;}}{y}_c^{\&prime;}+{\mathrm{my}}_{o^{\&prime;}}{x}_c^{\&prime;}\\ I_{\mathrm{xz}}={\mathrm{mx}}_{o^{\&prime;}}{z}_{o^{\&prime;}}+{\mathrm{mx}}_{o^{\&prime;}}{z}_c^{\&prime;}+{\mathrm{mz}}_{o^{\&prime;}}{x}_c^{\&prime;}\\ I_{\mathrm{yz}}={\mathrm{my}}_{o^{\&prime;}}{z}_{o^{\&prime;}}+{\mathrm{my}}_{o^{\&prime;}}{z}_c^{\&prime;}+{\mathrm{mz}}_{o^{\&prime;}}{y}_c^{\&prime;}\end{array}\right.---\left(4\right)$

3) as O " be the barycenter of rigid body, the principal axis of inertia that in O ' x ' y ' z ', each coordinate axis is rigid body simultaneously:

$\left\{\begin{array}{c}{I}_{\mathrm{xy}}={\mathrm{mx}}_{o^{\&prime;}}{y}_{o^{\&prime;}}\\ I_{\mathrm{xz}}={\mathrm{mx}}_{o^{\&prime;}}{z}_{o^{\&prime;}}\\ I_{\mathrm{yz}}={\mathrm{my}}_{o^{\&prime;}}{z}_{o^{\&prime;}}\end{array}\right.---\left(5\right)$

Three, the general type of product of inertia rotational transform

As shown in Figure 6, the initial point of coordinate system O ' x ' y ' z ' and Oxyz coincides, coordinate system O ' x ' y ' z ' with respect to the locus of Oxyz by 3 Eulerian angle φ that can independently convert, θ, determine.

Due in practice, it is the rotational transform of the product of inertia from O ' x ' y ' z ' to Oxyz that people consider many, so we discuss this rotational transform.

Coordinate system O ' x ' y ' z ' to transformation matrix of coordinates total between Oxyz is:

$\left\{\begin{array}{c}{I}_{\mathrm{xy}}=\frac{1}{2}\left(\sin 2\mathrm{\&phi;}\right)I_{x^{\&prime;2}}-\frac{1}{2}\left(\sin 2\mathrm{\&phi;}\right)I_{y^{\&prime;2}}+\left(\cos 2\mathrm{\&phi;}\right)I_{x^{\&prime;}{y}^{\&prime;}}\\ I_{\mathrm{xz}}=\left(\cos \mathrm{\&phi;}\right)I_{x^{\&prime;}{z}^{\&prime;}}-\left(\sin \mathrm{\&phi;}\right)I_{y^{\&prime;}{z}^{\&prime;}}\\ I_{\mathrm{yz}}=\left(\sin \mathrm{\&phi;}\right)I_{x^{\&prime;}{z}^{\&prime;}}+\left(\cos \mathrm{\&phi;}\right)I_{y^{\&prime;}{z}^{\&prime;}}\end{array}\right.---\left(7\right)$

Concurrent aces is the rotational transform rule of the product of inertia, right-hand man's rule: coincident axes is that thumb points to, and is tied to the coordinate system of amount to be asked by the coordinate of known quantity, uses formula (7) while rotating for the right hand.

Four, the calculating of the product of inertia of side arm and weight

Because of coordinate axis x ', y ', the principal axis of inertia that z ' is square-section, I _{x ' y '}=0, I _{x ' z '}=0, I _{y ' z '}=0, the product of inertia that can be obtained rotating later square-section by formula (7) is:

$\left\{\begin{array}{c}{I}_{\mathrm{xy}}=\frac{1}{2}\left(\sin 2\mathrm{\&alpha;}\right)I_{y^{\&prime;2}}-\frac{1}{2}\left(\sin 2\mathrm{\&alpha;}\right)I_{x^{\&prime;2}}\\ I_{\mathrm{xz}}=0\\ I_{\mathrm{yz}}=0\end{array}\right.---\left(8\right)$

By formula (1), can be obtained, square-section, in the process around the rotation of x axle, except centrifugal force is to the square of z axle, makes in addition the moment to z axle of cross section self rotation, and does not have moment to produce to other two axles.

For square-section its substitution formula (8) is obtained to the postrotational product of inertia in square-section is:

$\left\{\begin{array}{c}{I}_{\mathrm{xy}}=\frac{1}{2}\sin 2\mathrm{\&alpha;}(-\frac{m}{12}{b}^2)\\ I_{\mathrm{xz}}=0\\ I_{\mathrm{yz}}=0\end{array}\right.---\left(9\right)$

After trying to achieve the product of inertia in cross section, utilize formula (1), can obtain the moment of square-section to z, then along whole side arm integration, obtain the added force of inertia square producing in side arm rotary course:

$\left\{\begin{array}{c}{M}_{\mathrm{Ix}}=0\\ M_{\mathrm{Iy}}=\mathrm{\&rho;}\frac{1}{2}\sin 2\mathrm{\&alpha;}(-\frac{{\mathrm{ab}}^3}{12}){\mathrm{L\&omega;}}^2\\ M_{\mathrm{Iz}}=0\end{array}\right.---\left(10\right)$

Second step, sets up the parametrization computation model of asymmetric double arm configuration

$\begin{array}{c}{F}_Y=\frac{(h^2-a^2){\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_1}{2}+(h-\frac{d}{2}){\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_{2}R+(h-\frac{H}{2}){\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_{3}B\\ F_Z=-\frac{1}{2}{\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_2(R^2+\mathrm{RD})\sin \mathrm{\&alpha;}-\frac{1}{2}{\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_3(B^2+2\mathrm{RB}+\mathrm{BD})\sin \mathrm{\&alpha;}\\ M_X=\frac{1}{2}(R^2+\mathrm{RD})a{\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_2\sin \mathrm{\&alpha;}+\frac{1}{2}(B^2+\mathrm{BD}+2\mathrm{BR})a{\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_3\sin \mathrm{\&alpha;}\\ M_Y=\frac{1}{6}(\frac{3D^{2}R}{4}+\frac{3D{R}^2}{2}+R^3){\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_2\sin 2\mathrm{\&alpha;}\\ +\frac{1}{6}(B^3+\frac{3B^{2}D}{2}+3B^{2}R+\frac{2{\mathrm{BD}}^2}{4}+3\mathrm{BDR}+3{\mathrm{BR}}^2){\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_3\sin 2\mathrm{\&alpha;}\\ m_Z=\frac{1}{2}(h-d/2)(R^2+\mathrm{DR}){\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_2\cos \mathrm{\&alpha;}\\ +\frac{1}{2}(h-H/2)(B^2+\mathrm{DB}+2\mathrm{RB}){\mathrm{\&omega;}}^2\mathrm{\&rho;}{S}_3\cos \mathrm{\&alpha;}\end{array}---\left(11\right)$

The 3rd step, solves the system of equations of foundation

Can find out that above system of equations is for owing to determine Nonlinear System of Equations, the variable that meets above system of equations has infinite a plurality of.For asking above solution of equations, can only carry out step by step, getting some of them variable is definite value, asks its dependent variable.Utilize the fsolve function in MATLAB to solve, code programs.Because variable is more, establishment loop program solves, and obtains many groups and separates, and seeks most suitable one group in many groups solution of gained.Design proposal schematic diagram as shown in Figure 7 to 10.

Design example

Example 1: if when the load working condition of design is (12),

F _X＝120kN

F _Y＝979kN

F _Z＝360kN (12)

M _X＝-168kN·m

M _Y＝5kN·m

M _Z＝-188kN·m

Adopt single armed weight model to be difficult to reach requirement, can adopt both arms weight, design size is (13)

R1＝0.772m，B1＝0.230m，R2＝0.767m，B2＝0.480m

S1＝0.2m ²，S2＝0.072m ²，S3＝0.175m ²，S4＝0.0353m ² (13)

S5＝0.087m ²，α＝15°，β＝102°

Under this design proposal, the load that the both arms charger of designing produces is:

F _X＝0kN

F _Y＝979.369kN

F _Z＝187.121kN (14)

M _X＝-167.473kN·m

M _Y＝4.975kN·m

M _Z＝-187.908kN·m

Table 1 expectation value and design load must compare

	F x/KN	F y/KN	F z/KN	M x/KNm	M y/KNm	M z/KNm
							Expectation value	120	979	360	-168	5	-188
Design load	0	979.36	187.12	-167.473	4.975	-187.908
							Error (%)	-	0.036772	-	0.31369	0.5	0.04894

Example 2: if when the load working condition of design is (15).

F _X＝300kN

F _Y＝1129kN

F _Z＝375kN (15)

M _X＝-184kN·m

M _Y＝11kN·m

M _Z＝-230kN·m

Can adopt both arms weight, design size is:

R1＝0.650m，B1＝0.250m，R2＝0.650m，B2＝0.5m

S1＝0.2m ²，S2＝0.105m ²，S3＝0.2m ²，S4＝0.06m ² (16)

S5＝0.12m ²，α＝5°，β＝85°

Under this design proposal, the load that the both arms charger of designing produces is:

F _X＝0kN

F _Y＝1129.590kN

F _Z＝206.094kN (17)

M _X＝-184.454kN·m

M _Y＝11.350kN·m

M _Z＝-230.266kN·m

Table 2 expectation value and design load must compare

	F x/KN	F y/KN	F z/KN	M x/KNm	M y/KNm	M z/KNm
							Expectation value	300	1129	375	-184	11	-230.

Design load	0	1129.590	206.094	-184.454	11.350	-230.266
							Error (%)	-	0.052259	-	0.246739	3.181818	0.115652

From the design result error analysis of above-mentioned example 1 and example 2, adopt asymmetric both arms centrifugal loading device of the present invention, can accurately simulate the suffered centrifugal force F of (average error < 1%) propeller hub _ywith thrust moment M _x, torsion moment M _y, roll adjustment rotating vane moment of torsion M _zthree moments.

Claims (3)

1. an asymmetric both arms centrifugal loading device, comprise two side arms that arrange on a column being connected with propeller hub, column, each side arm end respectively has a weight, it is characterized in that: described two side arms and weight above thereof by column when propeller hub rotating shaft is rotated, the torsion moment of flexure and the thrust moment of flexure opposite direction that produce, rotating vane moment is identical with centrifugal force direction, adjust described two side arms length and with the angle of propeller hub rotating shaft, produce the suffered centrifugal force of accurate simulation propeller hub and thrust moment of flexure, torsion moment of flexure, three moments of roll adjustment rotating vane moment of torsion.

2. asymmetric both arms centrifugal loading device according to claim 1, is characterized in that: the length of described two side arms and cross section are all not identical, and the axis of two side arms is not identical with propeller hub rotating shaft angulation.

3. asymmetric both arms centrifugal loading device according to claim 1 and 2, is characterized in that: the length of described two side arms and with the angle parameterisable adjustment respectively of propeller hub rotating shaft.