CN104123458A - Transection type oblique crack rotor variable stiffness characteristic calculation method based on strain energy theory - Google Patents

Abstract

The invention discloses a transection type oblique crack rotor variable stiffness characteristic calculation method based on a strain energy theory. The transection type oblique crack rotor variable stiffness characteristic calculation method is characterized by comprising the following steps that (1) elastic strain energy of a crack-free rotor shaft unit is calculated, (2) displacement of the crack-free rotor shaft unit is calculated, (3) three types of crack rotor shaft stress density factors are calculated, (4) rotor shaft unit external elastic strain energy caused by cracks is calculated, (5) rotor shaft unit external displacement caused by the cracks is calculated, (6) flexibility coefficients of a crack rotor shaft unit are calculated, and (7) the variable stiffness characteristic is solved based on static balance transformation. The transection type oblique crack rotor variable stiffness characteristic calculation method has the advantages of being reasonable in design and accurate in calculation.

Description

A kind of transecting type shear crack rotor based on strain energy theory becomes stiffness characteristics computing method

Technical field

The present invention is based on strain energy theory, propose a kind of transecting type shear crack rotor and become stiffness characteristics computing method.Armature spindle adopts respectively Euler's beam and the beam element modeling of the pungent section of hophornbeam, and has considered degree of freedom longitudinal, crooked and torsion six direction.According to the stiffness matrix of calculation of flexibility factor crack element, and softness factor is tried to achieve according to the strain energy theory of crack element.These computing method have been established theoretical foundation for analyzing with the nonlinear characteristic that discloses transecting type shear crack vibration of rotor system.

Background technology

Crackle is a kind of important rotor-support-foundation system fault, if monitored not in time, will lead to disastrous consequence.In modern large rotating machinery equipment, rotor-support-foundation system works under harsh thermal stress and mechanical stress operating mode conventionally, and crackle often frequently occurs, so earns widespread respect and study about crackle modeling and troubleshooting issue.Existing crack model can be divided into two large classes, i.e. normal Crack ^[1]with breath crack ^[2].Normal Crack, by causing a local asymmetry of armature spindle rigidity to be often worth weakening, is applicable to the situation that tension acts on crack surface all the time.In breath crack situation, tension and compressive stress temporal evolution alternating action are on crack surface ^[3], can observe crackle present constantly open, closed breathing form, thereby cause that the cycle of armature spindle rigidity changes.At present, about crackle modeling and fault diagnosis research are mostly for the simplest lateral surfaces crackle ^[4-8], this has limited the practical application of achievement in research.Due to the complex nature of the problem, known shear crack research is fewer.Darpe has set up the rotor finite element model that comprises shear crack, and derives a new softness factor matrix based on fracturing mechanics, comprises the applied stress density factor of being introduced by direction of check in this matrix.And then, contrasted shear crack stiffness parameters and the coupled vibrations response characteristic different from transverse cracking rotor ^[9].Sekhar etc. have contrasted the model characteristics of shear crack and transverse cracking rotor, and the crackle of different depth, diverse location has been carried out to diagnosis research ^[10].

Formation of crack has multiple, such as transcrystalline cracking, mechanical fatigue crackle, the crackle that confluxes, grain-boundary crack, heat fatigue cracking, quenching crack, be full of cracks etc., produces reason very complicated.Theoretical according to failure analysis, its initiation and propagation direction of different formation of crack and different materials and shape is all not identical.What Darpe and Sekhar studied is vertical-type shear crack, and crack surface is perpendicular to xoy plane, as shown in Figure 2.

In fact, the shear crack of another kind of type is also very important for failure analysis, i.e. transecting type shear crack, crack surfaces transversal perpendicular to armature spindle axis but with xoy face out of plumb, as shown in Figure 3.

Summary of the invention

To the object of the invention is in order addressing the above problem, to have designed a kind of transecting type shear crack rotor based on strain energy theory and become stiffness characteristics computing method.

Realizing above-mentioned purpose technical scheme of the present invention is, a kind of transecting type shear crack rotor based on strain energy theory becomes stiffness characteristics computing method, it is characterized in that, the method comprises the steps:

1) flawless armature spindle unitary elasticity strain energy is calculated;

2) displacement of flawless armature spindle unit is calculated;

3) three types cracked rotor axial stress density factor is calculated;

4) the additional elastic strain energy in armature spindle unit that crackle causes is calculated;

5) the additional displacement in armature spindle unit that crackle causes is calculated;

6) cracked rotor axle unit calculation of flexibility factor;

7) the change stiffness characteristics based on static equilibrium conversion solves.

Described flawless armature spindle unitary elasticity strain energy calculating formula is:

m in formula ₁, M ₂for moment of flexure, T is moment of torsion, and F is axial force.G is rigidity modulus, and E is Young's modulus of elasticity.I is crackle cross sectional moment of inertia, I ₀for the crackle cross section utmost point moments of inertia.

The displacement meter formula of described flawless armature spindle unit is:

$\begin{array}{c}{u}_1^0{|}_E=\frac{{\∂U}^0}{{\∂P}_1}=\frac{P_{1}L}{\mathrm{AE}},u_2^0{|}_E=\frac{{\∂U}^0}{{\∂P}_2}=\frac{P_2{L}^3}{3\mathrm{EI}}-\frac{P_6{L}^2}{2\mathrm{EI}},u_3^0{|}_E=\frac{{\∂U}^0}{{\∂P}_3}=\frac{P_3{L}^3}{3\mathrm{EI}}+\frac{P_5{L}^2}{2\mathrm{EI}},\\ u_4^0{|}_E=\frac{{\∂U}^0}{{\∂P}_4}=\frac{P_{4}L}{{\mathrm{GI}}_0},u_5^0{|}_E=\frac{{\∂U}^0}{{\∂P}_5}=\frac{P_{5}L}{\mathrm{EI}}+\frac{P_3{L}^2}{2\mathrm{EI}},u_6^0{|}_E=\frac{{\∂U}^0}{\∂P_6}=\frac{P_{6}L}{\mathrm{EI}}-\frac{P_2{L}^2}{2\mathrm{EI}}.\end{array}.$

Described three types cracked rotor axial stress density factor calculating formula is:

Open Mode S/F:

$\begin{array}{c}{K}_1^I=\frac{P_1}{{\mathrm{pR}}^2}{\sin }^{2}q\sqrt{\mathrm{pa}}{F}_1(a/h),\\ K_2^I\frac{{\mathrm{kP}}_2}{{\mathrm{pR}}^2}\sin 2q\sqrt{\mathrm{pa}}{F}_1(a/h),\\ K_4^I=\frac{{2P}_4}{{\mathrm{pR}}^4}\sqrt{R^2-b^2}\sin 2q\sqrt{\mathrm{pa}}{F}_1(a/h),\\ K_5^I=\frac{4(P_5+P_3(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq}))}{{\mathrm{pR}}^4}b{\sin }^{2}q\sqrt{\mathrm{pa}}{F}_2(a/h),\\ K_6^I=\frac{4(P_2(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq})-P_6)}{{\mathrm{pR}}^4}\sqrt{R^2-b^2}{\sin }^{2}q\sqrt{\mathrm{pa}}{F}_2(a/h),\\ K_3^I=0.\end{array}$

Slippage Mode S/F:

$\begin{array}{c}{K}_1^{\mathrm{II}}=\frac{P_1}{{\mathrm{pR}}^2}\sin q\cos q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_2^{\mathrm{II}}=\frac{{\mathrm{kP}}_2}{{\mathrm{pR}}^2}\cos 2q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_4^{\mathrm{II}}=\frac{2P_4\sqrt{R^2-b^2}}{{\mathrm{pR}}^4}\cos 2q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\end{array}$

$\begin{array}{c}{K}_5^{\mathrm{II}}=\frac{4(P_5+P_3(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq}))}{{\mathrm{pR}}^4}b\sin q\cos q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_6^{\mathrm{II}}=\frac{4(P_2(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq})-P_6)}{{\mathrm{pR}}^4}\sqrt{R^2-b^2}\sin q\cos q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_3^{\mathrm{II}}=0\end{array}$

Shear mode SIF:

$\begin{array}{c}{K}_3^{\mathrm{III}}=\frac{{\mathrm{kp}}_3}{{\mathrm{pR}}^2}\sin q\sqrt{\mathrm{pa}}{F}_{\mathrm{III}}(a/h),\\ K_4^{\mathrm{III}}=\frac{{2P}_4}{{\mathrm{pR}}^4}b\sin q\sqrt{\mathrm{pa}}{F}_{\mathrm{III}}(a/h),\\ K_1^{\mathrm{III}}=K_2^{\mathrm{III}}=K_5^{\mathrm{III}}=K_6^{\mathrm{III}}=0.\end{array}$

Wherein

$\begin{array}{c}{F}_1=\sqrt{\frac{2h}{\mathrm{\π\α}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)\frac{0.752+2.02(\mathrm{\α}/h)+0.37{[1-\sin (\mathrm{\π\α}/2h)]}^3}{\cos (\mathrm{\π\α}/2h)}}\\ F_2=\sqrt{\frac{2h}{\mathrm{\π\α}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)\frac{0.923+0.199{[1-\sin (\mathrm{\π\α}/2h)]}^4}{\cos (\mathrm{\π\α}/2h)}}\\ F_{\mathrm{II}}=\frac{1.122-0.561(\mathrm{\α}/h)+0.085{(\mathrm{\α}/h)}^2+0.18{(\mathrm{\α}/h)}^3}{\sqrt{1-(\mathrm{\α}/h)}}\\ F_{\mathrm{III}}=\sqrt{\frac{2h}{\mathrm{\π\α}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)}\end{array}.$

The additional elastic strain energy calculating formula in armature spindle unit that described crackle causes is:

$J\left(A\right){|}_E=\frac{1}{E^{\text{'}}}[{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{Ii}}{|}_E\right)}^2+{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{IIi}}{|}_E\right)}^2+\mathrm{\η}{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{IIIi}}{|}_E\right)}^2]\·,$ E '=E/ in formula (1-v) and η=1+v.V is Poisson ratio, K _ii| _efor crack displacement opens the stress intensity factor of pattern, K _iIi| _efor the stress intensity factor of crack displacement slippage pattern, K _iIIi| _efor the stress intensity factor of crack displacement shear mode, i=1 ..., 6.

The additional displacement meter formula in armature spindle unit that crackle causes is:

j in formula (A) | _ethe strain energy density function providing according to fracturing mechanics concept during for the modeling of employing Euler beam element, is tried to achieve by the right formula described in 5 of will going.

Described cracked rotor axle unit softness factor g _ij| _ecalculating formula is:

The described change stiffness characteristics based on static equilibrium conversion solves formula into [K] ^c| _e=[T] G| _e[T] ^t, G| in formula _efor flexibility matrix, by [G] | _e={ g _ij| _e, i, j=1 ..., 6 provide, softness factor g _ij| _ecan calculate with reference to content described in claim 7, T is inversion matrix,

${\left[T\right]}^T=\left[\begin{array}{cccccccccccc}1& 0& 0& 0& 0& 0& -1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& -1& 0& 0& 0& l\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& 0& -l& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& -1& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& -1& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& -1\end{array}\right],$ [K] ^c| _efor the change stiffness matrix based on static equilibrium conversion.

Utilize the transecting type shear crack rotor based on strain energy theory that technical scheme of the present invention is made to become stiffness characteristics computing method, reasonable in design, result of calculation accuracy is high, and the transecting type shear crack rotor that can well calculate based on strain energy theory becomes stiffness characteristics.

Accompanying drawing explanation

Fig. 1 is the schematic flow sheet that the transecting type shear crack rotor based on strain energy theory of the present invention becomes stiffness characteristics computing method:

The stressed coordinate system of Fig. 2 vertical-type shear crack rotor and crack surface;

The stressed coordinate system of Fig. 3 lateral type shear crack rotor and crack surface.

Embodiment

Below in conjunction with accompanying drawing, the present invention is specifically described, if Fig. 1 is the schematic flow sheet that the transecting type shear crack rotor based on strain energy theory of the present invention becomes stiffness characteristics computing method, as shown in the figure, armature spindle adopts respectively two kinds of Euler's beam element and the beam element modelings of the pungent section of hophornbeam, and considered longitudinally, the crooked and degree of freedom of reversing six direction as Fig. 3 be one section of armature spindle that comprises transecting type shear crack, rotating disk quality is m, and rotating shaft diameter is D, and length is L.Crackle is positioned at left end x place, and crack depth is a.Unit stressing conditions is as follows: shearing P ₂, P ₃with P ₈, P ₉, moment of flexure P ₅, P ₆with P ₁₁, P ₁₂, axial force P ₁with P ₇, moment of torsion P ₄with P ₁₀.

The technical program be take Euler's beam element modeling armature spindle and is set forth the process solve the change stiffness characteristics based on static equilibrium conversion as example, its basic resolution principle is: according to the stiffness matrix of calculation of flexibility factor crack element, and softness factor is tried to achieve according to the Castingliano strain energy theory of crack element,

$u_i=\frac{\∂U}{{\∂P}_i}=\frac{{\∂U}^0}{{\∂P}_i}+\frac{{\∂U}^c}{{\∂P}_i}=u_i^0+u_i^c$

U in formula ⁰for the strain energy of flawless armature spindle, U ^cthe additional strain energy causing for crackle.U _iwith P _ibe respectively displacement and power along i coordinate direction.

Embodiment 1

Transecting type shear crack rotor based on the modeling of Euler's beam element becomes stiffness characteristics and calculates

While adopting the modeling of Euler's beam element, ignore detrusion on crack surface, only considered axial force, moment of torsion and Moment.The elastic strain energy U of flawless armature spindle unit in this case ₀| _ecan be expressed as

$U^0{|}_E=\frac{1}{2}\∫[\frac{M_1^2}{\mathrm{EI}}+\frac{M_2^2}{\mathrm{EI}}+\frac{T^2}{{\mathrm{GI}}_0}+\frac{F^2}{\mathrm{AE}}]\mathrm{dx}$

M in formula ₁, M ₂for moment of flexure, T is moment of torsion, and F is axial force.G is rigidity modulus, and E is Young's modulus of elasticity.I is crackle cross sectional moment of inertia, I ₀for the crackle cross section utmost point moments of inertia.

By Fig. 2, there is M ₁=P ₂x-P ₆, M ₂=P ₃x+P ₅, T=P ₄and F=P ₁.Thereby formula (2) becomes

$U^0{|}_E=\frac{1}{2}[\frac{P_1^{2}L}{\mathrm{AE}}+\frac{P_2^2{L}^3}{3\mathrm{EI}}+\frac{P_3^2{L}^3}{3\mathrm{EI}}+\frac{P_4^{2}L}{{\mathrm{GI}}_0}+\frac{P_5^{2}L}{\mathrm{EI}}+\frac{P_6^{2}L}{\mathrm{EI}}-\frac{P_2{P}_6{L}^2}{\mathrm{EI}}+\frac{P_3{P}_5{L}^2}{\mathrm{EI}}]$

The displacement of flawless Euler beam element for

$\begin{array}{c}{u}_1^0{|}_E=\frac{{\∂U}^0}{{\∂P}_1}=\frac{P_{1}L}{\mathrm{AE}},u_2^0{|}_E=\frac{{\∂U}^0}{{\∂P}_2}=\frac{P_2{L}^3}{3\mathrm{EI}}-\frac{P_6{L}^2}{2\mathrm{EI}},u_3^0{|}_E=\frac{{\∂U}^0}{{\∂P}_3}=\frac{P_3{L}^3}{3\mathrm{EI}}+\frac{P_5{L}^2}{2\mathrm{EI}},\\ u_4^0{|}_E=\frac{{\∂U}^0}{{\∂P}_4}=\frac{P_{4}L}{{\mathrm{GI}}_0},u_5^0{|}_E=\frac{{\∂U}^0}{{\∂P}_5}=\frac{P_{5}L}{\mathrm{EI}}+\frac{P_3{L}^2}{2\mathrm{EI}},u_6^0{|}_E=\frac{{\∂U}^0}{{\∂P}_6}=\frac{P_{6}L}{\mathrm{EI}}-\frac{P_2{L}^2}{2\mathrm{EI}}.\end{array}$

The additional displacement that crackle causes is

U ^c| _E＝∫ _AJ(A)| _EdA.

J in formula (A) | _ethe strain energy density function providing according to fracturing mechanics concept during for the modeling of employing Euler beam element, is provided by following formula

$J\left(A\right){|}_E=\frac{1}{E^{\′}}[{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{Ii}}{|}_E\right)}^2+{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{IIi}}{|}_E\right)}^2+\mathrm{\η}{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{IIIi}}{|}_E\right)}^2].$

E '=E/ in formula (1-v) and η=1+v.V is Poisson ratio, J _ii| _efor crack displacement opens the stress intensity factor of pattern, K _iIi| _efor the stress intensity factor of crack displacement slippage pattern, K _iIIi| _efor the stress intensity factor of crack displacement shear mode, i=1 ..., 6.

These stress intensity factors (SIF) are provided by following formula:

(1) open Mode S IF:

$\begin{array}{c}{K}_1^I=\frac{P_1}{p{R}^2}{\sin }^{2}q\sqrt{\mathrm{pa}}{F}_1(a/h),\\ K_2^I=\frac{k{P}_2}{p{R}^2}\sin 2q\sqrt{\mathrm{ap}}{F}_1(a/h),\\ K_4^I=\frac{2P_4}{p{R}^4}\sqrt{R^2-b^2}\sin 2q\sqrt{\mathrm{pa}}{F}_1(a/h),\\ K_5^I=\frac{4(P_5+P_3(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq}))}{p{R}^4}b{\sin }^{2}q\sqrt{\mathrm{pa}}{F}_2(a/h),\\ K_6^I=\frac{4(P_2(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq})-P_6)}{p{R}^4}\sqrt{R^2-b^2}{\sin }^{2}q\sqrt{\mathrm{pa}}{F}_2(a/h),\\ K_3^I=0.\end{array}$

(2) slippage Mode S IF:

$\begin{array}{c}{K}_1^{\mathrm{II}}=\frac{P_1}{p{R}^2}\sin q\cos q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_2^{\mathrm{II}}=\frac{k{P}_2}{p{R}^2}\cos 2q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\end{array}$

$\begin{array}{c}{K}_4^{\mathrm{II}}\frac{{2P}_4\sqrt{R^2-b^2}}{p{R}^4}\cos 2q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_5^{\mathrm{II}}=\frac{4(P_5+P_3(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq}))}{{\mathrm{pR}}^4}b\sin q\cos q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_6^{\mathrm{II}}=\frac{4(P_2(x-(\sqrt{R^2-b^2}-0.5a)\mathrm{ctgq})-P_6)}{p{R}^4}\sqrt{R^2-b^2}\sin q\cos q\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_3^{\mathrm{II}}=0\end{array}$

(3) shear mode SIF:

$\begin{array}{c}{K}_3^{\mathrm{III}}=\frac{k{P}_3}{p{R}^2}\sin q\sqrt{\mathrm{pa}}{F}_{\mathrm{III}}(a/h),\\ K_4^{\mathrm{III}}=\frac{2P_4}{p{R}^4}b\sin q\sqrt{\mathrm{pa}}{F}_{\mathrm{III}}(a/h),\\ K_1^{\mathrm{III}}=K_2^{\mathrm{III}}=K_5^{\mathrm{III}}=K_6^{\mathrm{III}}=0.\end{array}$

Wherein

$\begin{array}{c}{F}_1=\sqrt{\frac{2h}{\mathrm{\π\α}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)\frac{0.752+2.02(\mathrm{\α}/h)+0.37{[1-\sin (\mathrm{\π\α}/2h)]}^3}{\cos (\mathrm{\π\α}/2h)}}\\ F_2=\sqrt{\frac{2h}{\mathrm{\π\α}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)\frac{0.923+0.199{[1-\sin (\mathrm{\π\α}/2h)]}^4}{\cos (\mathrm{\π\α}/2h)}}\\ F_{\mathrm{II}}=\frac{1.122-0.561(\mathrm{\α}/h)+0.085{(\mathrm{\α}/h)}^2+0.18{(\mathrm{\α}/h)}^3}{\sqrt{1-(\mathrm{\α}/h)}}\\ F_{\mathrm{III}}=\sqrt{\frac{2h}{\mathrm{\π\α}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)}\end{array}$

Apply the strain energy J (A) of these stress intensity factor (SIF) formula and formula (6) | _eexpression formula, tries to achieve the additional displacement components u that crackle causes _i ^c| _e, i=1 ..., 6.And then, global displacement u _i| _ecan be by adding flawless rotor unit displacement components u _i ⁰| _etry to achieve, can be write as matrix form u _i| _e=u _i ⁰| _e+ u _i ^c| _e.=[G] | _ep _i, G| wherein _efor flexibility matrix, by [G] | _e={ g _ij| _e, i, j=1 ..., 6 provide.Softness factor g _ij| _efor

Consider finite element static equilibrium condition, adopt following formula by conversion T, to solve stiffness matrix by flexibility matrix:

{q _1-12} ^T＝[T]{q _1-6} ^T，

In formula, transformation matrix is

${\left[T\right]}^T=\left[\begin{array}{cccccccccccc}1& 0& 0& 0& 0& 0& -1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& -1& 0& 0& 0& l\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& 0& -l& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& -1& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& -1& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& -1\end{array}\right]$

Thereby transecting type shear crack rotor Euler beam element stiffness matrix is

[K] ^c| _E＝[T]G| _E[T] ^T

Embodiment 2

Transecting type shear crack rotor based on the beam element modeling of the pungent section of hophornbeam becomes stiffness characteristics and calculates

For the pungent section of hophornbeam beam element, the detrusion on crack surface must take in, thus the elastic strain energy U of flawless armature spindle unit ⁰| _tcan be written as

$U^0{|}_T=\frac{1}{2}\∫[\frac{{\mathrm{\α}}_s{V}_1^2}{\mathrm{GA}}+\frac{{\mathrm{\α}}_s{V}_2^2}{\mathrm{GA}}+\frac{M_1^2}{\mathrm{EI}}+\frac{M_2^2}{\mathrm{EI}}+\frac{T^2}{{\mathrm{GI}}_0}+\frac{F^2}{\mathrm{AE}}]\mathrm{dx}$

V in formula ₁, V ₂for shearing force, and V ₁=P ₂, V ₂=P ₃.α _sfor the detrusion factor.

Flawless Euler beam element displacement components u _i ⁰| _tfor

$\begin{array}{c}{u}_1^0{|}_T=u_1^0{|}_E,u_4^0{|}_T=u_4^0{|}_E,u_5^0{|}_T=u_5^0{|}_E,u_6^0{|}_T=u_6^0{|}_E,\\ u_2^0{|}_T=\frac{\∂U^0{|}_T}{\∂P_2}=(\frac{{\mathrm{\α}}_{s}L}{\mathrm{GA}}+\frac{L^3}{3\mathrm{EI}})P_2-\frac{P_6{L}^2}{2\mathrm{EI}},\\ u_3^0{|}_T=\frac{\∂U^0{|}_T}{\∂P_3}=(\frac{{\mathrm{\α}}_{s}L}{\mathrm{GA}}+\frac{L^3}{3\mathrm{EI}})P_3+\frac{P_5{L}^2}{2\mathrm{EI}}.\end{array}$

When adopting respectively the pungent section of Euler's beam and hophornbeam beam element to carry out modeling, (SIF) is different for the stress intensity factor, and difference is provided by following formula:

K _iI2| _e=K _iII3| _e=0. Euler's beam element

$\left.\begin{array}{c}{K}_{\mathrm{II}2}{|}_T=\frac{k{P}_2}{P{R}^2}\sqrt{\mathrm{pa}}{F}_{\mathrm{II}}(a/h),\\ K_{\mathrm{III}3}{|}_T=\frac{k{P}_3}{p{R}^2}\sqrt{\mathrm{pa}}{F}_{\mathrm{III}}(a/h).\end{array}\right\}$ The pungent section of hophornbeam beam element

In formula, k is shearing distribution modifying factor.Other the stress intensity factor is the same, can be with reference to formula above.

Finally, under transecting type shear crack, hophornbeam pungent section beam element and the difference of Euler's beam element on softness factor are provided by following formula:

$\left.\begin{array}{c}{g}_{22}{|}_E=\frac{L^3}{3\mathrm{EI}}+{\∫}_A[\frac{{2k}^2\mathrm{\α}{F}_{\mathrm{II}}^2}{\mathrm{\πE}{R}^4}+\frac{{8x}^2{h}^2\mathrm{\α}{F}_2^2}{\mathrm{\πE}{R}^8}]\mathrm{dA},\\ g_{33}{|}_E=\frac{L^3}{3\mathrm{EI}}+{\∫}_A[\frac{{32x}^2\mathrm{\α}{\mathrm{\β}}^2{F_1}^2}{\mathrm{\πE}{R}^8}+\frac{2\mathrm{\η}{k}^2\mathrm{\α}{F}_{\mathrm{III}}^2}{\mathrm{\πE}{R}^4}]\mathrm{dA},\\ g_{24}{|}_E=g_{42}{|}_E=0,g_{34}{|}_E=g_{43}{|}_E=0.\end{array}\right\}$ Euler's beam element

$\left.\begin{array}{c}{g}_{22}{|}_T=\frac{{\mathrm{\α}}_{s}L}{\mathrm{GA}}+\frac{L^3}{3\mathrm{EI}}+{\∫}_A[\frac{{2k}^2\mathrm{\α}{F}_{\mathrm{II}}^2}{\mathrm{\πE}{R}^4}+\frac{{8x}^2{h}^2\mathrm{\α}{F}_2^2}{\mathrm{\πE}{R}^8}]\mathrm{dA},\\ g_{33}{|}_T=\frac{{\mathrm{\α}}_{s}L}{\mathrm{GA}}+\frac{L^3}{3\mathrm{EI}}+{\∫}_A[\frac{{32x}^2\mathrm{\α}{\mathrm{\β}}^2{F_1}^2}{\mathrm{\πE}{R}^8}+\frac{2\mathrm{\η}{k}^2\mathrm{\α}{F}_{\mathrm{III}}^2}{\mathrm{\πE}{R}^4}]\mathrm{dA},\\ g_{24}{|}_T=g_{42}{|}_T{\∫}_A\frac{4\mathrm{k\α\β}{F}_{\mathrm{II}}^2}{\mathrm{\πE}{R}^6}\mathrm{dA},\\ g_{34}{|}_T=g_{43}{|}_T={\∫}_A\frac{2\mathrm{\ηkh\α}{F}_{\mathrm{III}}^2}{\mathrm{\πE}{R}^6}\mathrm{dA}.\end{array}\right\}$ The pungent section of hophornbeam beam element

Other softness factor is the same, can be with reference to formula above.Calculate softness factor g _ij| _tafter, can calculate with reference to the formula in formula embodiment 1 stiffness matrix [K] of the pungent section of transecting type shear crack rotor hophornbeam beam element ^c| _t.

Technique scheme has only embodied the optimal technical scheme of technical solution of the present invention, and those skilled in the art have all embodied principle of the present invention to some changes that wherein some part may be made, within belonging to protection scope of the present invention.

Claims (8)

1. the transecting type shear crack rotor based on strain energy theory becomes stiffness characteristics computing method, it is characterized in that, the method comprises the steps:

1) flawless armature spindle unitary elasticity strain energy is calculated;

2) displacement of flawless armature spindle unit is calculated;

3) three types cracked rotor axial stress density factor is calculated;

4) the additional elastic strain energy in armature spindle unit that crackle causes is calculated;

5) the additional displacement in armature spindle unit that crackle causes is calculated;

6) cracked rotor axle unit calculation of flexibility factor;

7) the change stiffness characteristics based on static equilibrium conversion solves.

2. the transecting type shear crack rotor based on strain energy theory according to claim 1 becomes stiffness characteristics computing method, it is characterized in that, described flawless armature spindle unitary elasticity strain energy calculating formula is: $U^0{|}_E=\frac{1}{2}\∫[\frac{M_1^2}{\mathrm{EI}}+\frac{M_2^2}{\mathrm{EI}}+\frac{T^2}{{\mathrm{GI}}_0}+\frac{F^2}{\mathrm{AE}}]\mathrm{dx},$ M in formula ₁, M ₂for moment of flexure, T is moment of torsion, and F is axial force.G is rigidity modulus, and E is Young's modulus of elasticity./ be crackle cross sectional moment of inertia ,/ ₀for the crackle cross section utmost point moments of inertia.

3. the transecting type shear crack rotor based on strain energy theory according to claim 1 becomes stiffness characteristics computing method, it is characterized in that, the displacement meter formula of described flawless armature spindle unit is: $\begin{array}{c}{u}_1^0{|}_E=\frac{{\∂U}^0}{{\∂P}_1}=\frac{P_{1}L}{\mathrm{AE}},u_2^0{|}_E=\frac{{\∂U}^0}{{\∂P}_2}=\frac{P_2{L}^3}{3\mathrm{EI}}-\frac{P_6{L}^2}{2\mathrm{EI}},u_3^0{|}_E=\frac{{\∂U}^0}{{\∂P}_3}=\frac{P_3{L}^3}{3\mathrm{EI}}+\frac{P_5{L}^2}{2\mathrm{EI}},\\ u_4^0{|}_E=\frac{{\∂U}^0}{{\∂P}_4}=\frac{P_{4}L}{{\mathrm{GI}}_0},u_5^0{|}_E=\frac{{\∂U}^0}{{\∂P}_5}=\frac{P_{5}L}{\mathrm{EI}}+\frac{P_3{L}^2}{2\mathrm{EI}},u_6^0{|}_E=\frac{{\∂U}^0}{\∂P_6}=\frac{P_{6}L}{\mathrm{EI}}-\frac{P_2{L}^2}{2\mathrm{EI}}.\end{array}.$

4. the transecting type shear crack rotor based on strain energy theory according to claim 1 becomes stiffness characteristics computing method, it is characterized in that, described three types cracked rotor axial stress density factor calculating formula is:

Open Mode S/F:

Slippage Mode S/F:

Shear mode SIF:

$\begin{array}{c}{K}_3^{\mathrm{III}}=\frac{{\mathrm{kP}}_3}{p{R}^2}\sin q\sqrt{\mathrm{pa}}{F}_{\mathrm{III}}(a/h),\\ K_4^{\mathrm{III}}=\frac{2P_4}{p{R}^4}b\sin q\sqrt{\mathrm{pa}}{F}_{\mathrm{III}}(a/h),\\ K_1^{\mathrm{III}}=K_2^{\mathrm{III}}=K_5^{\mathrm{III}}=K_6^{\mathrm{III}}=0.\end{array}$

Wherein

$\begin{array}{c}{F}_1=\sqrt{\frac{2h}{\mathrm{\π\α}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)\frac{0.752+2.02(\mathrm{\α}/h)+0.37{[1-\sin (\mathrm{\π\α}/2h)]}^3}{\cos (\mathrm{\π\α}/2h)}}\\ F_2=\sqrt{\frac{2h}{\mathrm{\π\α}}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)\frac{0.923+0.199{[1-\sin (\mathrm{\π\α}/2h)]}^4}{\cos (\mathrm{\π\α}/2h)}\\ F_{\mathrm{II}}=\frac{1.122-0.561(\mathrm{\α}/h)+0.085{(\mathrm{\α}/h)}^2+0.18{(\mathrm{\α}/h)}^3}{\sqrt{1-(\mathrm{\α}/h)}}\\ F_{\mathrm{III}}=\sqrt{\frac{2h}{\mathrm{\π\α}}}\tan \left(\frac{\mathrm{\π\α}}{2h}\right)\end{array}.$

5. the transecting type shear crack rotor based on strain energy theory according to claim 1 becomes stiffness characteristics computing method, it is characterized in that, the additional elastic strain energy calculating formula in armature spindle unit that described crackle causes is: $J\left(A\right){|}_E=\frac{1}{E\′}[{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{Ii}}{|}_E\right)}^2+{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{IIi}}{|}_E\right)}^2+\mathrm{\η}{\left(\underset{i=1}{\overset{6}{\mathrm{\Σ}}}{K}_{\mathrm{IIIi}}{|}_E\right)}^2]\·,$ E '=E/ in formula (1-v) and η=1+v.V is Poisson ratio, K _ii| _efor crack displacement opens the stress intensity factor of pattern, K _iIi| _efor the stress intensity factor of crack displacement slippage pattern, K _iIIi| _efor the stress intensity factor of crack displacement shear mode, i=1 ..., 6.

6. the transecting type shear crack rotor based on strain energy theory according to claim 1 becomes stiffness characteristics computing method, it is characterized in that, the additional displacement meter formula in armature spindle unit that crackle causes is: $u_i^c{|}_E=\frac{{\∂U}^c{|}_E}{{\∂P}_i},U^c{|}_E={\∫}_{A}J\left(A\right){|}_E\mathrm{dA}.,$ J in formula (A) | _ethe strain energy density function providing according to fracturing mechanics concept during for the modeling of employing Euler beam element, is tried to achieve by the right formula described in 5 of will going.

7. the transecting type shear crack rotor based on strain energy theory according to claim 1 becomes stiffness characteristics computing method, it is characterized in that described cracked rotor axle unit softness factor g _ij| _ecalculating formula is:

8. transecting type shear crack rotor based on strain energy theory according to claim 1 becomes stiffness characteristics computing method, it is characterized in that, the described change stiffness characteristics based on static equilibrium conversion solves formula into [K] ^c| _e=[T] G| _e[T] ^t, G| in formula _efor flexibility matrix, by [G] | _e={ g _ij| _e, i, j=1 ..., 6 provide, softness factor g _ij| _ecan calculate with reference to content described in claim 7, T is inversion matrix,

${\left[T\right]}^T=\left[\begin{array}{cccccccccccc}1& 0& 0& 0& 0& 0& -1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0& 0& -1& 0& 0& 0& l\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& 0& -l& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& -1& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& -1& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& -1\end{array}\right],$ [K] ^c| _efor the change stiffness matrix based on static equilibrium conversion.