CN201883348U - Double-inclined weld face structure of seamless rail high bearing capability and without bump and shake during operation - Google Patents

Abstract

The utility model relates to a double-inclined weld face structure of a seamless rail with high bearing capability and without bump and shake during operation, which belongs to the technical field of welding of rail train seamless rails. A weld joint of a rail adopts a double-inclined weld face, an angle alpha is formed between the double-inclined weld face and the vertical direction of the rail, an angle beta is formed between the double-inclined weld face and the horizontal direction of the rail, the double-inclined weld faces of two parallel rails are arranged in a staggered manner, and the staggered length is greater than the length of a compartment. The double-inclined weld face structure can further improve the stress state of the rail weld face, increase the bearing capacity of the weld face and simultaneously eliminate the vertical bump and horizontal shake of a train; and the double-inclined weld faces of the two parallel rails are arranged in a horizontally staggered manner, and the staggered length is greater than the length of a compartment, so that the operation stability and durability of the train can be enhanced, and the conventional simple and convenient exothermic welding technique can be used for seamless rail welding.

Description

Operation does not have the big gapless rail diclinic solder side structure of vibrations supporting capacity of jolting

Technical field

The utility model belongs to the technical field of track train gapless rail welding, is to change the existing axial monocline solder side of rail that is diagonal to into the diclinic solder side specifically.

Background technology

The inventor's patent application case (application number is 200910206270.7) and (application number is 201010250990.6), the both has increased the tangential and axial carrying capacity of solder side.The former has designed parallel with the y axle, with the monocline solder side at x axle oblique α angle, when wheel tread passes through the weld seam of rail head tread monocline solder side, can only eliminate pitching of train; The latter has designed parallel with the x axle, monocline solder side with y axle oblique β angle, when the weld seam of wheel rim by rail head side monocline solder side, can only eliminate train about vibrations, still there is deficiency in this stability, safety and durability to the high-speed heavy-load trains operation.Therefore, design and a kind ofly can increase the tangential and axial carrying capacity of solder side, and when the weld seam of train by solder side, can eliminate simultaneously again pitch and about the diclinic solder side dimensional orientation structure of vibrations, be the problem that solution is needed in the welding of rail integral seamless badly.

Summary of the invention

The utility model is in order to solve the problem of above-mentioned existence, a kind of tangential and axial carrying capacity that can increase solder side is proposed, can when the weld seam of train by solder side, can eliminate simultaneously again pitch with about shake, and can use the diclinic solder side new construction of exothermic welding technology effectively.

Below in conjunction with accompanying drawing and subordinate list the utility model is described:

The utility model is made of rail and weld seam, and wherein the weld seam of rail 1 is diclinic solder side A _{α β}, diclinic solder side A _{α β}With the dimensional orientation pass of rail 1 be: the straight plane of ABCD is the cross section A perpendicular to the z axle ₀, the straight plane of ABCD obtains the ABEG inclined-plane behind y axle rotation alpha angle, and the ABEG inclined-plane obtains BEDH inclined-plane, i.e. diclinic solder side A again after BE rotates the β angle _{α β}Diclinic solder side A _{α β}Become the α angle with the x axle, become the β angle with Y-axis.

In subordinate list 1, subordinate list 2 and the subordinate list 3: when

When

When

The collocation value at corresponding α angle and β angle all can be chosen to be α=30 °, β=30 °, α=30 °, β=45 °, α=45 °, β=30 °, α=45 °, β=45 °, α=45 °, β=60 °, α=60 °, β=45 °, α=60 °, β=30 °, α=30 °, β=60 °, α=60 °, β=60 °

Above σ _0zFor along z to normal stress allowable, τ _0yAnd τ _0xBe respectively along y to x to shear stress allowable.

Wheel tread and wheel rim and rail be for to contact synchronously, i.e. wheel tread 6 and diclinic solder side A _{α β}Overlap joint about the inclined weld 5 of rail head tread is, corresponding wheel rim 8 and diclinic solder side A _{α β}Inclined weld 7 in the rail head side is the front and back overlap joint.

Article two, the diclinic solder side A on the parallel steel rails 1 _{α β}Dislocation is arranged, and dislocation length is greater than the length in a joint compartment.

Diclinic solder side A _{α β}The welding procedure that adopts is exothermic welding.

Further specify below in conjunction with accompanying drawing:

The position relation of diclinic solder side as shown in Figure 1, A _{α β}For become the α angle to become the diclinic solder side at β angle with vertical y with horizontal x.

The formation of diclinic solder side: diclinic solder side A _{α β}Position relation formation as shown in Figure 2, solder side ABCD is the cross section A perpendicular to the z axle ₀, the ABCD plane obtains the ABEG inclined-plane behind y axle rotation alpha angle, and the ABEG inclined-plane obtains inclined-plane BEDH, i.e. double inclined plane A again after BE rotates the β angle then _{α β}N is the normal vector of diclinic solder side BEDH, it respectively with the x axle the angle of cut is (π-α '), with the angle of cut of y axle and z axle be β ', γ ' angle.Wherein:

α＝∠CBE，β＝∠CDF，α′＝∠OCB，β′＝∠OCD，γ′＝∠OCE；

2. bibeveled trigonometric function relation

With reference to Fig. 2:

CF＝BC·sinα，CE＝CF/cosα，BF＝BC·cosα，OF＝CF·sinβ＝BC·sinα·sinβ，

EF＝CF·tgα＝BC·sinα·tgα，OC＝CF·cosβ＝BC·sinα·cosβ

$\∠\mathrm{BOF}=\mathrm{arctg}\frac{\mathrm{BF}}{\mathrm{OF}}=\mathrm{arctg}\frac{\mathrm{BC}\·\cos \mathrm{\α}}{\mathrm{BC}\·\sin \mathrm{\α}\·\sin \mathrm{\β}}=\mathrm{arctg}\frac{\cos \mathrm{\α}}{\sin \mathrm{\α}\·\sin \mathrm{\β}},$

$\∠\mathrm{EOF}=\mathrm{arctg}\frac{\mathrm{BF}}{\mathrm{OF}}=\mathrm{arctg}\frac{\mathrm{BC}\·\cos \mathrm{\α}\·\mathrm{tg\α}}{\mathrm{BC}\·\sin \mathrm{\α}\·\sin \mathrm{\β}}=\mathrm{arctg}\frac{\mathrm{tg\α}}{\sin \mathrm{\β}}.$

By on can get

$\cos (\mathrm{\π}-{\mathrm{\α}}^{\′})=-\cos {\mathrm{\α}}^{\′}=-\frac{\mathrm{OC}}{\mathrm{BC}}=-\sin \mathrm{\α}\cos \mathrm{\β}---\left(1\right)$

$\sin {\mathrm{\&alpha;}}^{\&prime;}=\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2{\mathrm{\&alpha;}}^{\&prime;}}=\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&alpha;}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}---\left(2\right)$

$\cos {\mathrm{\&beta;}}^{\&prime;}=\cos (\frac{\mathrm{\&pi;}}{2}-\mathrm{\&beta;})=\sin \mathrm{\&beta;}---\left(3\right)$

$\sin {\mathrm{\&beta;}}^{\&prime;}=\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2{\mathrm{\&beta;}}^{\&prime;}}=\sqrt{1-{\sin }^2\mathrm{\&beta;}}=\cos \mathrm{\&beta;}---\left(4\right)$

$\cos {\mathrm{\&gamma;}}^{\&prime;}=\frac{\mathrm{OC}}{\mathrm{CE}}=\frac{\mathrm{CF}\&CenterDot;\cos \mathrm{\&beta;}}{\frac{\mathrm{CF}}{\cos \mathrm{\&alpha;}}}=\cos \mathrm{\&alpha;}\&CenterDot;\cos \mathrm{\&beta;}---\left(5\right)$

$\sin {\mathrm{\&gamma;}}^{\&prime;}=\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2{\mathrm{\&gamma;}}^{\&prime;}}=\sqrt{1-{\sin }^2\mathrm{\&alpha;}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}---\left(6\right)$

$\cos \&angle;\mathrm{BOF}=\frac{\sin \mathrm{\&alpha;}\sin \mathrm{\&beta;}}{\sqrt{1-{\sin }^2\mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}}---\left(7\right)$

$\sin \&angle;\mathrm{BOF}=\frac{\sin \mathrm{\&alpha;}\sin \mathrm{\&beta;}}{\sqrt{1-{\sin }^2\mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}}---\left(8\right)$

$\cos \&angle;\mathrm{EOF}=\frac{\sin \mathrm{\&alpha;}\sin \mathrm{\&beta;}}{\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}}---\left(9\right)$

$\sin \&angle;\mathrm{EOF}=\frac{\sin \mathrm{\&alpha;}}{\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}}---\left(10\right)$

3. the stress analysis of diclinic solder side

As shown in Figure 2, act on A along the train direction of advance _{α β}Pulling force on the face is F _z, train acts on A _{α β}Vertical load on the face is F _y, wheel rim acts on A _{α β}The transverse load of face is F _x, owing to act on A ₀The normal stress of face upper edge z direction is σ _0z=F _0z/ A ₀, act on A ₀The shear stress along x and y direction on the face is τ _0x=F _0x/ A ₀, τ _0y=F _0y/ A ₀, F _0x, F _0y, F _0zBe peak load, τ _0x, τ _0yBe A ₀Maximum shear stress on the face, σ _0zBe A ₀Maximum tension stress on the face.So F _0x=A ₀τ _0x, F _0y=A ₀τ _0y, F _0z=A ₀σ _0z, with τ _0x, τ _0y, σ _0zBe decided to be allowable stress, more help the reliability of analysis result.

Act on double inclined plane A _{α β}On stress be respectively f _x=F _x/ A _{α β}, f _y=F _y/ A _{α β}, f _z=F _z/ A _{α β}, by A ₀=A _{α β}Cos γ ' and cos γ '=cos α cos β can get

A ₀＝Aαβcosα·cosβ(11)

Because of F _z=F _0z, F _y=F _0y, F _x=F _0xSo, by Fig. 1 and (11) formula as can be known:

$f_z=\frac{{\mathrm{<figure-callout\; id="0"\; label="F\; z\; A"\; filenames="HSA00000298434200021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}{A_{}}\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;}={\mathrm{\&sigma;}}_{0z}\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;},$ $f_y=\frac{{\mathrm{<figure-callout\; id="0"\; label="F\; z\; A"\; filenames="HSA00000298434200021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}{A_{}}\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;}={\mathrm{\&tau;}}_{0y}\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;},$

$f_x=\frac{{\mathrm{<figure-callout\; id="0"\; label="F\; z\; A"\; filenames="HSA00000298434200021.png"\; state="\{\{state\}\}">F</figure-callout>}}_z}{A_{}}\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;}={\mathrm{\&tau;}}_{0y}\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;}.$

3.1 the combined stress of normal stress

With reference to Fig. 2, act on A _{α β}σ on the face _Zn=f _zCos γ ', σ _Xn=f _xCos α ' and σ _Yn=f _yCos β ' is normal stress.Therefore, act on A _{α β}Positive combined stress on the face is σ _h=σ _Zn+ σ _Yh-σ _XnSo can get

σ _h＝f _z·cosγ′+f _y·cosβ′-f _x·cosα′

After (1), (3), (5) formula substitution dissolved, just

σ _h＝σ _0z·cos ²αcos ²β+τ _0y·cosαsinβcosβ-τ _0x·sinαcosαcos ²β(12)

3.2 the combined stress of shear stress

As shown in Figure 2, f _x, f _y, f _zWith act on A _{α β}Shear stress be respectively τ _x, τ _y, τ _z, shear stress τ _x, τ _y, τ _zMust be orthogonal to normal n, and drop on A _{α β}On the face.Therefore, τ _x=f _xSin α ', τ _y=f _ySin β ', τ _z=f _zSin γ '.Again because of τ _x, τ _y, τ _zBe to drop on A _{α β}On the face.So τ _xWith BO in the same way, τ _yWith OF in the same way, τ _zWith EO in the same way.We are concerned about be x to shear stress and y to close shear stress, so need are with shear stress τ _zBe decomposed into x to shear stress τ _ZxWith y to shear stress τ _Zy, corresponding to the shear stress exploded view of Fig. 2 as shown in Figure 3, by sine,

$\frac{{\mathrm{\&tau;}}_z}{\sin \&angle;\mathrm{BOF}}=\frac{{\mathrm{\&tau;}}_{\mathrm{zx}}}{\sin \&angle;\mathrm{EOF}}=\frac{{\mathrm{\&tau;}}_{\mathrm{zy}}}{\sin (\mathrm{\&pi;}-\&angle;\mathrm{BOF}-\&angle;\mathrm{EOF})}$

So, τ _Zx=τ _zSin ∠ EOF/sin ∠ BOF; τ _Zy=τ _zCos ∠ EOF+ τ _zSin ∠ EOFctg ∠ BOF

So, along the combined stress τ of x direction shear stress _Xh

${\mathrm{\&tau;}}_{\mathrm{xh}}={\mathrm{\&tau;}}_x-{\mathrm{\&tau;}}_{\mathrm{zx}}={\mathrm{\&tau;}}_x-\frac{{\mathrm{\&tau;}}_z\sin \&angle;\mathrm{EOF}}{\sin \&angle;\mathrm{BOF}}---\left(13\right)$

With (8), (10) formula substitution (13) formula, after dissolving, just

${\mathrm{\&tau;}}_{\mathrm{xh}}=({\mathrm{\&tau;}}_{0x}\&CenterDot;\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;}-{\mathrm{\&sigma;}}_{0z}\&CenterDot;\sin \mathrm{\&alpha;}\cos \mathrm{\&beta;})\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}---\left(14\right)$

Combined stress τ along y direction shear stress _Yh

τ _yh＝τ _y-τ _zy＝τ _y-τ _z(cos∠EOF+sin∠EOFctg∠BOF)(15)

With (7)-(10) formula substitutions (15) formula, after dissolving, just

τ _yh＝τ _0y·cosαcos ²β-σ _0z·sinβcosβ(16)

3.3 reducing of diclinic solder side stress

Compare and act on A ₀Normal stress σ on the face _0zWith shear stress τ _0x, τ _0y, act on A _{α β}Normal stress σ on the face _hWith shear stress τ _Xh, τ _YhBe reduced to Δ σ and Δ τ _x, Δ τ _y, can try to achieve by (12) formula and (14) formula, (16) formula respectively:

$\mathrm{\&Delta;\&sigma;}=\frac{{\mathrm{\&sigma;}}_{0z}-{\mathrm{\&sigma;}}_n}{{\mathrm{\&sigma;}}_{0z}}=1-{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}-\frac{{\mathrm{\&tau;}}_{0y}}{{\mathrm{\&sigma;}}_{0z}}\cos \mathrm{\&alpha;}\sin \mathrm{\&beta;}\cos \mathrm{\&beta;}+\frac{{\mathrm{\&tau;}}_{0x}}{{\mathrm{\&sigma;}}_{0z}}\sin \mathrm{\&alpha;}\cos {\mathrm{\&beta;}\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}---\left(17\right)$

$\mathrm{\&Delta;}{\mathrm{\&tau;}}_x=\frac{{\mathrm{\&tau;}}_{0x}-{\mathrm{\&tau;}}_{\mathrm{xh}}}{{\mathrm{\&tau;}}_{0x}}=1-(\cos \mathrm{\&alpha;}\cos \mathrm{\&beta;}-\frac{{\mathrm{\&sigma;}}_{0z}}{{\mathrm{\&tau;}}_{0x}}\sin \mathrm{\&beta;}\cos \mathrm{\&beta;})\sqrt{1-{\mathrm{<figure-callout\; id="2"\; label="sin"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">sin</figure-callout>}}^2\mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}---\left(18\right)$

$\mathrm{\&Delta;}{\mathrm{\&tau;}}_y=\frac{{\mathrm{\&tau;}}_{0y}-{\mathrm{\&tau;}}_{\mathrm{yh}}}{{\mathrm{\&tau;}}_{0y}}=1-\cos \mathrm{\&alpha;}{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}+\frac{{\mathrm{\&sigma;}}_{0z}}{{\mathrm{\&tau;}}_{0y}}\sin \mathrm{\&beta;}\cos \mathrm{\&beta;}---\left(19\right)$

4. the increase of operation stability and supporting capacity

4.1 the stability of operation

In the train direction of advance, the relation that wheel acts on rail head tread and rail head side inclined weld simultaneously as shown in Figure 4.When train when becoming the oblique solder side weld seam at α angle with the x axle, wheel tread 6 is not to contact fully with rail head inclined weld 5, but contacts with two parts inside and outside the inclined weld, so a weight is just born jointly by inside and outside two parts of rail head.Become the inclined weld at β angle with the y axle when the train process, wheel rim 8 neither contact fully with rail head side inclined weld 7, but contact with two parts before and after the inclined weld, so the transverse load of wheel rim is just born jointly by two parts before and after the rail head side, this just fundamentally eliminated simultaneously since welding depression cause pitching of train and about vibrations.Therefore, this has just further increased the ride comfort and the stability of train operation than monocline solder side.

4.2 the increase of supporting capacity

Owing to act on A ₀Z on the face to, y to, x to peak load be F _0z, F _0y, F _0x, and F _0z=A ₀σ _0z, F _0y=A ₀τ _0y, F _0x=A ₀τ _0x, σ wherein _0z, τ _0y, τ _0xBe allowable stress.Act on A _{α β}The z of face is F to, x to peak load to, y _z, F _y, F _xSo, F _zAt A _{α β}Normal load on the face is F _{α β n}=F _zCos γ ', F _yAt A _{α β}Face along y to load be F _{α β y}=F _yCos β ', F _xAt A _{α β}Face upper edge x to load be F _{α β x}=F _xCos α ', and F _{α β n}=A _{α β}σ _{α β n}, F _{α β y}=A _{α β}τ _{α β y}, F _{α β x}=A _{α β}τ _{α β x}, use allowable stress σ _0z, τ _0y, τ _0xReplace σ _{α β n}, τ _{α β y}, τ _{α β x}The comparativity of not losing points and analysing is so have: F _{α β n}=σ _0zA _{α β}, F _{α β y}=τ _0yA _{α β}, F _{α β x}=τ _0xA _{α β}So: with respect to A ₀Maximum z on the face is to load F _0z, A _{α β}On the face z to supporting capacity increase to

${\mathrm{\&Delta;F}}_z=\frac{F_z-F_{0z}}{F_{0z}}=\frac{F_{\mathrm{\&alpha;\&beta;n}}/\cos {\mathrm{\&gamma;}}^{\&prime;}}{F_{0z}}-1=\frac{A_{\mathrm{\&alpha;\&beta;}}\&CenterDot;{\mathrm{\&sigma;}}_{0z}/\cos {\mathrm{\&gamma;}}^{\&prime;}}{A_0\&CenterDot;{\mathrm{\&sigma;}}_{0z}}-1=\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2{\mathrm{\&gamma;}}^{\&prime;}}---\left(20\right)$

With respect to A ₀Maximum y on the face is to load F _0y, A _{α β}Y increasing on the face to supporting capacity

${\mathrm{\&Delta;F}}_y=\frac{F_y-F_{0y}}{F_{0y}}=\frac{F_{\mathrm{\&alpha;\&beta;y}}/\cos {\mathrm{\&beta;}}^{\&prime;}}{F_{0y}}-1=\frac{1}{\cos {\mathrm{\&beta;}}^{\&prime;}\cos {\mathrm{\&gamma;}}^{\&prime;}}-1---\left(21\right)$

With respect to A ₀Maximum x on the face is to load F _0x, A _{α β}X increasing on the face to supporting capacity

$\mathrm{\&Delta;}{F}_x=\frac{F_x-F_{0x}}{F_{0x}}=\frac{1}{\cos {\mathrm{\&alpha;}}^{\&prime;}\cos {\mathrm{\&gamma;}}^{\&prime;}}-1---\left(22\right)$

With (1), (3), correspondingly substitution of (5) formula (20), (21), (22) formula, just

$\mathrm{\&Delta;}{F}_z=\frac{1}{{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&alpha;}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}-1---\left(23\right)$

$\mathrm{\&Delta;}{F}_y=\frac{1}{\cos \mathrm{\&alpha;}\&CenterDot;\sin \mathrm{\&beta;}\cos \mathrm{\&beta;}}-1---\left(24\right)$

$\mathrm{\&Delta;}{F}_x=\frac{1}{\sin \mathrm{\&alpha;}\cos \mathrm{\&alpha;}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="cos"\; filenames="HSA00000298434200011.png,HSA00000298434200032.png"\; state="\{\{state\}\}">cos</figure-callout>}}^2\mathrm{\&beta;}}-1---\left(25\right)$

The beneficial effects of the utility model are:

1. when the inclined weld of train by the diclinic solder side, can eliminate simultaneously pitch and about vibrations.

2. all reduce along the vertical and horizontal pure shear stress of rail diclinic solder side and along the pure tensile stress of the rail direction of motion.

3. horizontal, the vertical and axial supporting capacity along rail all increases.

4. mistake is arranged before and after the diclinic solder side of two parallel steel rails, more increased the safety of train operation, because the diclinic solder side has reduced pure tensile stress and pure shear stress, and increased horizontal, vertical and axial supporting capacity, and therefore adopt the exothermic welding welding, also can guarantee the reliability of diclinic solder side, this has just improved welding efficiency, simplify welding procedure, saved welding cost, and can factory's weldering or online welding.

5. be specially adapted to the welding of heavy haul train and EMU rail, also can be used for the gapless rail welding of city railway train.

In a word, the utility model can not only increase the supporting capacity of solder side, reduce the axial tension stress and the horizontal shear stress of solder side, and can eliminate simultaneously train pitch and about vibrations, exothermic welding effectively can also be used for the welding of gapless rail, the rail of railway construction is not had seam welding have important change, be particularly suitable for the integral seamless welding of heavy duty, bullet train rail.

Description of drawings

Fig. 1 is diclinic solder side A _{α β}The dimensional orientation schematic diagram

Fig. 2 is diclinic solder side A _{α β}Formation and stress state figure

Fig. 3 is diclinic solder side A _{α β}Shear stress figure

Fig. 4 passes through the operation stability schematic diagram of rail head side inclined weld by rail head tread inclined weld and wheel rim for wheel tread

Fig. 5 is the oblique solder side dislocation layout viewing of two parallel steel rails

Fig. 6 is for becoming the monocline face orientation diagram at α angle with the x axle

Fig. 7 is for becoming the monocline face orientation diagram at β angle with the y axle

Wherein: 1. rail 2. rail heads 3. webs of the rail 4. flanges of rail 5. diclinic solder side A _{α β}Inclined weld 6. wheel treads 7. diclinic solder side A at the rail head tread _{α β}Inclined weld 8. wheel rims 9. in the rail head side become the weld seam 12. of the monocline solder side at β angle to cross the wheel rim of monocline weld seam 11 with the wheel tread 11. that the x axle hands over the weld seam 10. of the monocline solder side at α angle to cross monocline weld seam 9 with the y axle

The specific embodiment

The utility model is made of rail and diclinic solder side, two parallel steel rails 1 and diclinic solder side A _{α β}Be that dislocation is arranged, dislocation length is greater than the length in a joint compartment, and welding manner is existing exothermic welding technology.

Below be the production instance of the utility model " operation does not have the big gapless rail diclinic solder side structure of vibrations supporting capacity of jolting ":

In actual applications, roadbed is that existing steel concrete does not have tiny fragments of stone, coal, etc. roadbed, rail is the rail of the used various models of existing heavy-duty freight car, bullet train and city railway train, and the sleeper that rail is connected with roadbed, backing plate, backing plate set bolt and fastener are constant fully.

Owing to after having determined bibeveled α angle and β angle, could determine A _{α β}Face, positive tensile stress reduce Δ σ, pure shear stress reduce Δ τ _xWith Δ τ _yUnder the prerequisite that adheres to specification, could determine double inclined plane A _{α β}α and the value of β.Again since loading in A _{α β}Contain undetermined in the analytic expression that the face upper stress reduces (17)～(19) formula

With Therefore, known

With

The prerequisite of value under, just can try to achieve Δ σ, Δ τ corresponding to the value of α, β _x, Δ τ _yValue.Because train is in operation, and acts on A _{α β}Face upper edge z to maximum trailed load F _zGreater than wheel along y to maximum positive voltage load F _y, F _yAct on A greater than wheel rim _{α β}Maximum transversal load F on the face _xSo, corresponding maximum stress σ _0z, τ _0xAnd τ _0yBetween have σ _0z＞τ _0y＞τ _0xRelation in view of the above, is set respectively

Afterwards, the stress of trying to achieve again corresponding to α, β reduces value, lists in subordinate list 1, subordinate list 2 and subordinate list 3.Corresponding to the increase value of α, β angle supporting capacity, list in subordinate list 4.

According to the actual conditions of dissimilar train operations, determine

With

Value, the stress according to designing requirement reduces to be worth Δ σ, Δ τ again _y, Δ τ _x, just can determine A _{α β}The design α angle of face and the value at β angle.The determined value that α angle and β angle have been arranged just can be sawn into A with band saw or toothless saw with the termination of two sections rail to be welded _{α β}Double inclined plane, alignment up and down and is reserved suitable gap again, with the exothermic welding welding, the welding of double inclined plane joint is promptly finished in, polishing protruding with pusher, heat treatment.

By subordinate list 1, subordinate list 2 and subordinate list 3 as can be known, shear stress Δ τ _xWith Δ τ _yReduce value all greater than 100%.As if this is pondered-over at first view.But if analyze shear stress Fig. 3, it is doubted and just releases: because of τ _ZxAnd τ _ZyBe τ _zAlong the branch shear stress of x and y, and its direction and τ _xAnd τ _yDirection is opposite.So work as τ _ZxWith τ _xAbsolute value when equating, τ _ZyWith τ _yAbsolute value when equating, Δ τ _xWith Δ τ _yThe value that reduces equal 100%; Work as τ _ZxAnd τ _ZyAbsolute value greater than τ _xAnd τ _yAbsolute value the time, Δ τ _xWith Δ τ _yReduce value just greater than 100%.But, do not allow Δ τ _xWith Δ τ _yReduce value greater than 200%.As seen from Figure 2, normal stress σ _ZnCan not be greater than σ _Xn+ σ _Yn, so the value that reduces of Δ σ can not be greater than 100%.

Claims (2)

1. one kind is moved and is not had the big gapless rail diclinic solder side structure of vibrations supporting capacity of jolting, and is made of rail and weld seam, it is characterized in that the weld seam of rail (1) is diclinic solder side A _{A β}, diclinic solder side A _{A β}With the dimensional orientation pass of rail (1) be: the straight plane of ABCD is the cross section A perpendicular to the z axle ₀, the straight plane of ABCD obtains the ABEG inclined-plane behind y axle rotation alpha angle, and the ABEG inclined-plane obtains BEDH inclined-plane, i.e. diclinic solder side A again after BE rotates the β angle _{A β}Diclinic solder side A _{A β}Become the α angle with the x axle, become the β angle with Y-axis; Wheel tread (6) and wheel rim (8) and rail be for to contact synchronously, i.e. wheel tread (6) and diclinic solder side A _{A β}Overlap joint about the inclined weld (5) of rail head tread is, corresponding wheel rim (8) and diclinic solder side A _{A β}Inclined weld (7) in the rail head side is the front and back overlap joint; When

When

When

The collocation value at corresponding α angle and β angle all can be chosen to be α=30 °, β=30 °, or α=30 °, β=45 °, or α=45 °, β=30 °, or α=45 °, β=45 °, or α=45 °, β=60 °, or α=60 °, β=45 °, or α=60 °, β=30 °, or α=30 °, β=60 °, or α=60 °, β=60 °

Above σ _0zFor along z to normal stress allowable, τ _0yAnd τ _{0 χ}Be respectively along y to x to shear stress allowable.

2. do not have the big gapless rail diclinic solder side structure of vibrations supporting capacity of jolting by the described operation of claim 1, it is characterized in that the diclinic solder side A on two parallel steel rails (1) _{A β}Dislocation is arranged, and dislocation length is greater than the length in a joint compartment.

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