CN108287957A - A kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety analytical method - Google Patents
A kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety analytical method Download PDFInfo
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Abstract
The invention discloses a kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety analytical methods, including step：Large cargo transports axis vehicle main beam stress static indeterminacy mechanical analysis；Variation rigidity cantilever beam any position degree of disturbing under single form force effect；Variation rigidity simply supported beam any position degree of disturbing under single form force effect；Vehicle frame any position amount of deflection under single form force effect；Ask vehicle frame any position shearing force, moment of flexure and amount of deflection；Vehicle frame maximum support and reinforcing beam setting.The advantage of the invention is that：Longitudinal beam each point institute bending moment, shearing and amount of deflection are accurately calculated, its safety is demonstrated and proposes cargo support setting and the method for frame strength reinforcement position and size under the support of calculated specific data.
Description
Technical field
The present invention relates to traffic safety technical field, more particularly to a kind of highway awkward and lengthy cargo transport trailer master is vertical
Safety beam analysis method.
Background technology
Awkward and lengthy cargo transport plays very important effect to national major project construction, is transported in awkward and lengthy cargo
Cheng Zhong needs the transportation safety for ensureing awkward and lengthy cargo, and whether trailer longitudinal beam intensity is reliably awkward and lengthy cargo after loading
The important content of transportation safety.
Traditional trailer longitudinal beam strength check methods are according to the position of the fulcrum provided in service manual --- bearing capacity is bent
Whether line finds out position of the fulcrum permitted bearing capacity, checked [1] less than permitted bearing capacity with actual support power.The branch point
Set --- two fulcrum forces that cracking, which is trailer manufacturer, to be arranged symmetrically with trailer in vertical, and two stress
Equal is condition, the permitted bearing capacity under calculated difference position of the fulcrum.This method calculates simply, with convenience, but if
Two position of the fulcrum asymmetry of cargo or when power is unequal or fulcrum quantity more than two, with the condition of this method foundation not phase
Meet, the method can not be used；And it can not also be adopted when the trailer number of axle trailer number of axle corresponding with curve in data differs
With, but in practical applications, both of which largely exists.
Another method is strength of materials Method of Checking, and in terms of the stressing conditions of trailer main longitudinal grider, trailer main longitudinal grider is mainly held
It is typical supporting beam beam-type member by the pressure that the upward support force of frame suspension and cargo fulcrum are downward.Due to big part goods
Amount of substance is big, appearance and size is big, often there is multiple cargo fulcrums, so, the stress of trailer main longitudinal grider belong to supporting beam it is quiet not
Determine mechanics problem.Static indeterminacy mechanics problem solving complexity, thus, in most of practical applications, it is assumed that each cargo fulcrum stress phase
Deng verifying vehicle frame main longitudinal grider intensity as condition using this equal power, it is clear that the result that such method obtains is unreliable.Meanwhile
In practical applications, it when longeron insufficient strength, needs to add reinforcing beam for longeron, longeron can be regarded as to a kind of edge after addition
The supporting beam of longitudinal cross-section the moment of inertia variation, this more increases the difficulty of longeron intensive analysis.
Static indeterminacy mechanical analysis to beam, document [2-3] use FInite Element, different modulus are had studied based on basis of sensitivity analysis
The internal force of truss；Document [4] provides the cosine function computational methods of general member structure modal displacement；Document [5] is directed to more bars
Junction problem has studied the operation method based on displacement method using the geometrical relationship supplement Coordinate deformation equation of Deformation Member；Text
It offers [6] and modal displacement is calculated based on time-vector method research；Document [7] using velocity projections method research calculating static determinacy and it is quiet not
Fixed pole system structure interior joint displacement；Document [8] is had studied using pure mathematics operation and establishes deformation geometry side in indeterminate truss
The analytic method of journey；Document [9-10] has studied indeterminate truss Coordinate deformation equation using differential analysis method.Document above is not examined
Consider the situation of the section-variable of beam.In the case of the section-variable for considering beam, document [12], which proposes, uses three moments euqation
The stress of beam is analyzed, and then calculates stress, moment of flexure suffered by trailer main longitudinal grider each point, this approach application has been arrived extension by document [1]
During vehicle longitudinal strength is checked, but the method needs to calculate respectively across the moment of flexure area of pictural surface under across internal loading effect, when calculating amount of deflection
Using the conversion bending moment diagram of trailer girder as the dummy load distribution map of imaginary beam in conjugate beam computational methods, find out with practical beam respectively by
Empty distributed load small area between corresponding two adjacent coordinates of force coordinate and position of centre of gravity, then section deformation amount is calculated,
Bending moment diagram is mostly irregular figure, and areal calculation gets up comparatively laborious, and this method solution procedure is more complex, and calculation amount is larger.
Bibliography：
[1] Xiao builds limited liability company of English road heavy transportations technology [M] People's Transportation Press, and 2015:74-78；
[2] different modulus truss Direct And Inverse Problems of the Zhang Xiaoyue based on basis of sensitivity analysis solve the Dalian [D]:Dalian University of Technology
Engineering mechanics system, 2008:1-35；
[3] Yang Haitian, Zhang Xiaoyue, the numerical analysis of what suitable machine with draw tool truss problems of the modest based on basis of sensitivity analysis
[J] Computational Mechanics journals, 2011,28 (2):237-242；
[4] discussion [J] Xi'nan College of Forestry journals of the general member structure joint displacement calculation methods of the suitable loyalty of Yao, 2002,
22(1):61-63；
[5] Chen Ping, Chen Guoliang, general solution [J] the mechanics of the lower multi-rod converging problem of the general load effects of Yang Xu and practice,
2014,36(3):348-350；
[6] a kind of straightforward procedure [J] mechanics that displacement of joint in the osmanthus Feng Xian calculates and practice, 2002,24 (1):49-50；
[7] skill is applied in the Shanghai a kind of computational methods [J] of Zhu Yide static determinacy and static indeterminacy member structure interior joint displacement
Art institute journal, 2007,7 (1):33-35；
[8] Ni Er has analytic method [J] Anshan Iron ＆ Steel College journals that Geometric equation of deformation is established in indeterminate trusses,
1991,14(3):56-59；
[9] Bian Wenfeng, Dong are just building new method [J] Computational Mechanics journals of indeterminate truss Coordinate deformation equations, and 2002,
19(2):250-252；
[10] high Jinhua derives static indeterminacy truss Coordinate deformation equation [J] mechanics and puts into practice using differential method, and 1998,
20(5):67-68；
[11] Beijing Fan Qinshan, Wang Jing the mechanics of materials [M]:China Railway Press, 2016.06:139-140；
[12] Liu Guang good continuous beam three moments euqation is proved, understanding [J] engineering mechanics of expression formula and its decomposition it is miscellaneous
Will, 2003:262-265.
Invention content
The present invention in view of the drawbacks of the prior art, provides a kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety point
Analysis method, can effectively solve the problem that the above-mentioned problems of the prior art.
In order to realize the above goal of the invention, the technical solution adopted by the present invention is as follows：
A kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety analytical method, includes the following steps：
Step 1, large cargo transports axis vehicle main beam stress static indeterminacy mechanical analysis；
With vehicle goods generally research object, if each hydraulic axis support force is respectively：Fj(j=1,2 ..., N),
If axis vehicle supported at three point marshalling using 2 points in preceding, number of axle I, a little rear, the number of axle J, I+J=N are then built
Equation group is to find out each hydraulic axis support force under Liru：
Goods weight is G in formula, and center of gravity of goods is L apart from vehicle frame front end distanceg, vehicle frame weight is Gf, vehicle frame length is L.
Each wheelbase axis vehicle of axis vehicle front end distance is Xj(j=1,2 ..., N), N are the hydraulic pressure support number of axle.
If supported at three point using any preceding, number of axle I, 2 points in rear, number of axle J, calculation formula is identical.
Step 2, variation rigidity cantilever beam any position degree of disturbing under single form force effect；
If x is cantilever beam length, EI is cantilever beam bending stiffness；
Unit concentrfated load amount of deflection f caused by free end, rotational angle theta calculation formula are respectively：
Specific torque amount of deflection f caused by free end, rotational angle theta calculation formula are respectively：
It is respectively in free end amount of deflection f, rotational angle theta calculation formula caused by unit even distributed force：
For variation rigidity cantilever beam, if x indicates that, away from fixing end distance on cantilever beam, y indicates that concentrated force acts on cantilever beam
Distance of the position away from fixing end, if array z [0~H] be each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam z [i-1] with
It is EI [i] that z [i] section, which corresponds to rigidity, then under unit concentrfated load effect, away from the rotational angle theta at fixing end distance x on cantilever beamFx
Following formula calculating can be used：
If x is located at variation rigidity, cantilever beam kth section has if x≤y：
Wherein：
θ′FiIt is opposite this section of beginning corner in i-th section of cantilever beam end, has：θ′Fi=θ 'Fi(Mi)+θ′Fi(F), θ 'Fi(Mi)
It is end caused by i-th section of cantilever beam tail end section moment of flexure with respect to this section of beginning corner,
θ′Fi(F) it is with respect to this section of end beginning corner caused by i-th section of cantilever beam tail end section concentrated force,
θ′FxFor opposite kth section cantilever beam beginning corner at x,
Have
If x>Y has：θFx=θFy；
Under unit concentrfated load effect, away from the amount of deflection f at fixing end distance x on cantilever beamFxFollowing formula calculation can be used
Go out：
If x is located at variation rigidity, cantilever beam kth section has if x≤y：
Wherein：
f′FiIt is opposite this section of beginning amount of deflection in i-th section of cantilever beam end, has：
f′Fi(Mi) it is with respect to this section of end beginning amount of deflection caused by i-th section of cantilever beam tail end section moment of flexure,f′Fi(F) it is that end caused by i-th section of cantilever beam tail end section concentrated force relatively should
Section beginning amount of deflection, For i sections of amounts of deflection caused by i-th section of beginning corner.
f′FxFor opposite kth section cantilever beam beginning amount of deflection at x, have For opposite k sections of beginning amounts of deflection caused by section turn moment at x,
f′Fx(F) it is opposite k sections of beginning amounts of deflection caused by the concentrated force of section at x, For amount of deflection at x caused by kth section beginning corner.
If x>Y has：fFx=fFy+θFy(x-y)。
Variation rigidity cantilever beam is by Uniform Loads, cantilever beam length l, then under unit Uniform Loads, cantilever
Away from the rotational angle theta at fixing end distance x on beamqx, obtained using following formula：
Wherein, θ 'qiIt is opposite this section of beginning corner in i-th section of cantilever beam end, has：θ′qi=θ 'qi(q)+θ′qi(Mi)+θ′qi
(F), θ 'qi(q) it is with respect to this section of end beginning corner caused by i-th section of cantilever beam end even distributed force,θ′qi(Mi) with respect to this section of end beginning caused by i-th section of cantilever beam tail end section moment of flexure turns
Angle,θ′qi(F) end caused by i-th section of cantilever beam tail end section concentrated force is opposite should
Section beginning corner,
Wherein, θ 'qxFor opposite k sections of beginning corners at x, have：θ′qx=θ 'qx(q)+θ′qx(Mi)+θ′qx(F), θ 'qx(q) it is
Opposite k sections of beginning corners caused by opposite k sections of beginning even distributed forces at x,θ′qx(Mi) it is to be cut at x
Opposite k sections of beginning corners at x caused by the moment of flexure of face,θ′qx(F) it is that section is concentrated at x
Opposite k sections of beginning corners at x caused by power,
Under unit Uniform Loads, away from the amount of deflection f at fixing end distance x on cantilever beamqxIt is obtained using following formula：
Wherein, f 'qiIt is opposite this section of beginning amount of deflection in i-th section of cantilever beam end, has： f′qi(q) it is end caused by i-th section of cantilever beam end even distributed force
Opposite this section of beginning amount of deflection,f′qi(Mi) end caused by i-th section of cantilever beam tail end section moment of flexure
Opposite this section of beginning amount of deflection,f′qi(F) i-th section of cantilever beam tail end section concentrated force draws
With respect to this section of the end beginning amount of deflection risen, Begin for i-th section
Hold i sections of amounts of deflection caused by corner.
f′qxFor opposite kth section cantilever beam beginning amount of deflection at x, have f′qx(q) it is opposite k sections of beginning amounts of deflection caused by even distributed force at x,f′Fx(Mx) it is opposite k sections of beginning amounts of deflection caused by section turn moment at x,
f′Fx(F) it is opposite k sections of beginning amounts of deflection caused by the concentrated force of section at x, For amount of deflection at x caused by kth section beginning corner.
Step 3, variation rigidity simply supported beam any position degree of disturbing under single form force effect；
When variation rigidity simply supported beam being asked to deform, simply supported beam is converted to cantilever beam, a bearing of simply supported beam is converted into fixation
End, another bearing are replaced with restraining force, and there are certain rotational angle thetas with former simply supported beam position for cantilever beamA, value is by beam in restraining force
It is that zero condition is calculated to hold amount of deflection.
Under unit concentrfated load effect, y indicates that concentrated force acts on distance of the simply supported beam position away from its left end, and l indicates letter
Distance between two fulcrums of strutbeam, if array z [0~H] is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam z [i-1]
Rigidity corresponding with z [i] sections is EI [i], then：
Then under unit concentrfated load effect, the amount of deflection w at the x of simply supported beam any positionF(x, y) is obtained by following formula：
Under unit concentrfated load effect, the corner γ at the x of simply supported beam any positionF(x, y) is obtained by following formula：
Under unit Uniform Loads, l indicates that the distance between two fulcrums of simply supported beam, L indicate simply supported beam overall length, array z
[0~M] is each section of rigidity initial position of variation rigidity cantilever beam, and cantilever beam z [i-1] rigidity corresponding with z [i] sections is EI [i], then：
Then under unit Uniform Loads, the amount of deflection w at the x of simply supported beam any positionq(x, L, l) is obtained by following formula：
Under unit Uniform Loads, the corner γ at the x of simply supported beam any positionq(x, L, l) is obtained by following formula：
Step 4, vehicle frame any position amount of deflection under single form force effect；
Using vehicle frame as research object, if the power that cargo fulcrum is applied to vehicle frame is
Ni(i=1,2 ..., M), M count for cargo branch, and each cargo fulcrum is Y away from axis vehicle front end distancei(i=1,
2,…,M)。
If array s [0~P] is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam s [i-1] is corresponding with s [i] sections
Rigidity is GD [i]；
The power that each cargo fulcrum is applied to vehicle frame is found out, because large cargo cross sectional moment of inertia is far beyond vehicle frame the moment of inertia,
It is therefore assumed that cargo is rigid body with cargo support, the influence of the deformation of consideration vehicle frame main longitudinal grider then makees the branched point of cargo
It is reduced to using cargo fulcrum as bearing with the force structure of lower trailer main longitudinal grider, using trailer main longitudinal grider as beam, with crossbeam to main vertical
Beam action power and the stress sketch that vehicle frame its own gravity is load；
The superfluous constraint of cargo fulcrum at the 2nd to M-1 is released, statically indeterminate beam is made to become statically determinate beam；
With fFAmount of deflection caused by (x, y) indicates to act on the unit concentrated force of vehicle frame y location at the x of vehicle frame position, for
Vehicle frame any position degree of disturbing, can be used Load superposition and is calculated, that is, calculate separately each power amount of deflection caused by same position,
These amounts of deflection and be the final amount of deflection in the position.
The even distributed force formed to the weight of vehicle frame itself acts on vehicle frame situation, decomposes three kinds of stressing conditions, and vehicle frame is arbitrary
The sum of amount of deflection caused by position amount of deflection should be three kinds of stressing conditions in the position:
That is unit even distributed force amount of deflection caused by x at an arbitrary position：
fQ(x)=fQ1(x)+fQ2(x)+fQ3(x)
Wherein, fQ(x) amount of deflection caused by indicating unit even distributed force at x, fQ1(x)、fQ2(x)、fQ3(x) indicate single respectively
Position even distributed force caused amount of deflection at x.
Step 5, each cargo fulcrum is asked to be applied to the power N of vehicle framei；
Be zero for deformation compatibility condition with the amount of deflection at superfluous constraint, in conjunction be with joint efforts zero with to arbitrary point torque and be zero
Equilibrium condition then has：
In formula,It indicates
The power N of vehicle frame is applied to by cargo fulcrumi(i=1,2 ..., M), each hydraulic axis support force Fj(j=1,2 ..., N) and vehicle frame
The even distributed force that itself weight is formed at the superfluous constraint of statically indeterminate beam caused by amount of deflection be zero, M-2 can be constructed with this
A equation, to which equation group (2) can solve the power N that each cargo fulcrum is applied to vehicle framei(i=1,2 ..., M).
Step 6, vehicle frame any position shearing force, moment of flexure and amount of deflection are asked；
If vehicle frame is Q (x) by shearing for institute at x away from axis vehicle front end distance, then have：
Wherein：
If vehicle frame is M (x) for the interfaces x institute bending moment away from axis vehicle front end distance, then have：
If vehicle frame is that amount of deflection is ω at x away from axis vehicle front end distancex, then have：
Step 7, vehicle frame maximum support and reinforcing beam setting；
To the convex situation of longeron, the method for increasing the salient point movement upward of holder or two side stands is taken to reduce convex deformation；
To the recessed situation of longeron, takes plus pre- arch reduces concave deformation, steps are as follows：
71, two outermost end position of the fulcrum are set, the most outer segment fulcrum of cargo two should ensure that there is certain distance apart from cargo both ends,
If the distance is ldlAnd ldr；
72, each fulcrum stress is calculated, in the fulcrum of both ends, if certain fulcrum stress is negative and there are certain fulcrum previous steps to be
Fulcrum is added, then cancels the added fulcrum of previous step, fulcrum stress is recalculated, in next step in record revocation position；
73, vehicle frame amount of deflection everywhere is calculated, if somewhere amount of deflection is just and to be more than maximum allowable positive amount of deflection Δ laIf do not exist herein
Position queue is cancelled, and the position is more than away from closest pivot distance between fulcrum allows distance la, then fulcrum is added at this,
72 are gone to, otherwise, in next step；
74, vehicle frame amount of deflection everywhere is calculated, is positive value in each deflection value and is more than maximum allowable positive amount of deflection Δ laIf the position
Closest two pivot distances are set more than permission distance l between fulcruma, then closest two fulcrum is towards position movement Δ l, when mobile
Two pivot distances should be not less than la；
75, vehicle frame amount of deflection everywhere is calculated, if somewhere amount of deflection is negative and more than the maximum allowable negative amount of deflection Δ l of regulationb, then at this
Place adds pre- arch, predomed height to be equal to amount of deflection and maximum allowable negative amount of deflection Δ l at this in sectionbThe absolute value of difference.
Further, when the longitudinal strength deficiency of trailer, it is strong to improve longeron to install longitudinally reinforced girder steel additional on trailer
Degree, the reinforcing beam and trailer main longitudinal grider of installation can only be stacked together, and two beams have respective respectively in bending deformation process
Neutral axis；Under normal circumstances, since trailer beam is very long, deforming latter two beam still can be close together, thus can be reinforcing beam
It is seen integrally with girder, it is assumed that the moment of inertia increases Δ I after trailer girder adds reinforcing beam, the method for judging whether to add reinforcing beam
It is as follows：
Vehicle frame moment of flexure everywhere is calculated, since longeron one end, is added successively in the longeron section setting that moment of flexure is more than moment of flexure allowable
Brutal：If certain section of moment of flexure is more than to allow moment of flexure, reinforcing beam is set in the section, trailer girder adds increased inertia after reinforcing beam
Square is：((this section of maximal bending moment-moment of flexure allowable)/moment of flexure allowable) * safety factor * trailer girder the moment of inertia * trailer girders are allowable
Stress and reinforcing beam allowable stress ratio.Reinforcing beam both ends should at least exceed this section of both ends certain distance Δ lc。
Compared with prior art the advantage of the invention is that：With superfluous constraint point displacement be zero it is that condition establishes compatibility of deformation
Equation, and then find out the power that each holder under static indeterminacy stressing conditions acts on trailer girder；It is counted in longitudinal beam deflectometer, it will
Longitudinal beam simply supported beam stress check calculation is cantilever beam stress, and longitudinal beam is acquired based on the piecewise tempering addition method and Load superposition
Each point amount of deflection；Based on longeron force analysis, longitudinal beam each point institute bending moment, shearing and amount of deflection have been accurately calculated, has been verified
Its safety simultaneously proposes under the support of calculated specific data cargo support setting and frame strength reinforces position
And the method for size.
Description of the drawings
Fig. 1 is large cargo Stowage Plane；
Fig. 2 is vehicle goods Overall Analysis figure of the embodiment of the present invention；
Fig. 3 is that cantilever beam of embodiment of the present invention free end is acted on schematic diagram by concentrfated load；
Fig. 4 is cantilever beam of embodiment of the present invention free end by moment loading schematic diagram；
Fig. 5 is cantilever beam of the embodiment of the present invention by Uniform Loads schematic diagram；
Fig. 6 is that variation rigidity of embodiment of the present invention cantilever beam is acted on schematic diagram by concentrfated load；
Fig. 7 is variation rigidity of embodiment of the present invention cantilever beam by Uniform Loads schematic diagram；
Fig. 8 is that simply supported beam of the embodiment of the present invention is converted to cantilever beam schematic diagram；
Fig. 9 is vehicle frame force analysis figure of the embodiment of the present invention；
Figure 10 is vehicle frame stress sketch of the embodiment of the present invention；
Figure 11 is that the embodiment of the present invention releases superfluous constraint stress sketch；
Figure 12 is that even distributed force of the embodiment of the present invention acts on 1 exploded view of vehicle frame situation；
Figure 13 is that even distributed force of the embodiment of the present invention acts on 2 exploded view of vehicle frame situation；
Figure 14 is that even distributed force of the embodiment of the present invention acts on 3 exploded view of vehicle frame situation.
Specific implementation mode
To make the objectives, technical solutions, and advantages of the present invention more comprehensible, develop simultaneously embodiment referring to the drawings, right
The present invention is described in further details.
A kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety analytical method, includes the following steps：
Step 1, large cargo transports axis vehicle main beam stress static indeterminacy mechanical analysis；
Large cargo usually selects axis vehicle combined traffic, typical highway large cargo transport to load as shown in Figure 1.Figure
In, goods weight G, center of gravity of goods is L apart from vehicle frame front end distanceg, vehicle frame weight is Gf, vehicle frame length is L.Axis vehicle is each
Wheelbase axis vehicle front end distance is Xj(j=1,2 ..., N), N are the hydraulic pressure support number of axle.
With vehicle goods generally research object, if each hydraulic axis support force is respectively：Fj(j=1,2 ..., N), stress letter
Figure is as shown in Figure 2.
If axis vehicle supported at three point marshalling can then be established such as using (number of axle I) before 2 points, a little rear (number of axle J), I+J=N
Lower equation group is to find out each hydraulic axis support force (equal with group support force)：
If supported at three point uses before a bit (number of axle I), after 2 points (number of axle J), calculation formula is identical.
Step 2, variation rigidity cantilever beam any position degree of disturbing under single form force effect；
Reciprocity rigidity uniform cantilever beam, Fig. 3 indicate that cantilever beam free end is acted on by concentrfated load, and x is length of cantilever
Degree, EI is cantilever beam bending stiffness, and unit concentrfated load amount of deflection caused by free end, corner calculation formula are respectively：
Fig. 4 indicates cantilever beam free end by moment loading, and x is cantilever beam length, and EI is cantilever beam bending stiffness, unit
Torque amount of deflection caused by free end, corner calculation formula are respectively：
Fig. 5 indicates cantilever beam by Uniform Loads, and x is cantilever beam length, and EI is cantilever beam bending stiffness, and unit is equal
It is respectively in free end amount of deflection, corner calculation formula caused by cloth power：
For variation rigidity cantilever beam, Fig. 6 indicates that variation rigidity cantilever beam is acted on by concentrfated load, wherein x indicates cantilever beam
On away from fixing end distance, y indicates that concentrated force acts on distance of the cantilever beam position away from fixing end, if array z [0~H] is to become rigid
Each section of rigidity initial position of cantilever beam is spent, cantilever beam z [i-1] rigidity corresponding with z [i] sections is EI [i], then in unit concentrfated load
Under effect, away from the rotational angle theta at fixing end distance x on cantilever beamFxFollowing formula calculating can be used：
If x is located at variation rigidity, cantilever beam kth section has if x≤y：
Wherein：
θ′FiIt is opposite this section of beginning corner in i-th section of cantilever beam end, has：θ′Fi=θ 'Fi(Mi)+θ′Fi(F), θ 'Fi(Mi)
It is end caused by i-th section of cantilever beam tail end section moment of flexure with respect to this section of beginning corner,
θ′Fi(F) it is with respect to this section of end beginning corner caused by i-th section of cantilever beam tail end section concentrated force,
θ′FxFor opposite kth section cantilever beam beginning corner at x, have
If x>Y has：θFx=θFy。
Under unit concentrfated load effect, away from the amount of deflection f at fixing end distance x on cantilever beamFxFollowing formula calculation can be used
Go out：
If x is located at variation rigidity, cantilever beam kth section has if x≤y：
Wherein：
f′FiIt is opposite this section of beginning amount of deflection in i-th section of cantilever beam end, has： f′Fi(Mi) it is that end caused by i-th section of cantilever beam tail end section moment of flexure relatively should
Section beginning amount of deflection,f′Fi(F) caused by being i-th section of cantilever beam tail end section concentrated force
With respect to this section of end beginning amount of deflection, Cause for i-th section of beginning corner
I section amounts of deflection.
f′FxFor opposite kth section cantilever beam beginning amount of deflection at x, have f′Fx(Mx) it is opposite k sections of beginning amounts of deflection caused by section turn moment at x,
f′Fx(F) it is opposite k sections of beginning amounts of deflection caused by the concentrated force of section at x, For amount of deflection at x caused by kth section beginning corner.
If x>Y has：fFx=fFy+θFy(x-y)。
Fig. 7 indicates variation rigidity cantilever beam by Uniform Loads, cantilever beam length l, then in unit Uniform Loads
Under, away from the rotational angle theta at fixing end distance x on cantilever beamqxFollowing formula can be used to obtain：
Wherein, θ 'qiIt is opposite this section of beginning corner in i-th section of cantilever beam end, has：θ′qi=θ 'qi(q)+θ′qi(Mi)+θ′qi
(F), θ 'qi(q) it is with respect to this section of end beginning corner caused by i-th section of cantilever beam end even distributed force,
θ′qi(Mi) with respect to this section of end beginning corner caused by i-th section of cantilever beam tail end section moment of flexure,θ′qi(F) end caused by i-th section of cantilever beam tail end section concentrated force is begun with respect to the section
Corner is held,
Wherein, θ 'qxFor opposite k sections of beginning corners at x, have：θ′qx=θ 'qx(q)+θ′qx(Mi)+θ′qx(F), θ 'qx(q) it is
Opposite k sections of beginning corners caused by opposite k sections of beginning even distributed forces at x,θ′qx(Mi) it is to be cut at x
Opposite k sections of beginning corners at x caused by the moment of flexure of face,θ′qx(F) it is that section is concentrated at x
Opposite k sections of beginning corners at x caused by power,
Under unit Uniform Loads, away from the amount of deflection f at fixing end distance x on cantilever beamqxFollowing formula can be used to obtain
Go out：
Wherein, f 'qiIt is opposite this section of beginning amount of deflection in i-th section of cantilever beam end, has： f′q ′i(q) it is end caused by i-th section of cantilever beam end even distributed force
Opposite this section of beginning amount of deflection in end,f′qi(Mi) end caused by i-th section of cantilever beam tail end section moment of flexure
Opposite this section of beginning amount of deflection in end,f′qi(F) i-th section of cantilever beam tail end section concentrated force
With respect to this section of caused end beginning amount of deflection,It is i-th
I sections of amounts of deflection caused by section beginning corner.
f′qxFor opposite kth section cantilever beam beginning amount of deflection at x, have f′qx(q) it is opposite k sections of beginning amounts of deflection caused by even distributed force at x,f′Fx(Mx) it is opposite k sections of beginning amounts of deflection caused by section turn moment at x,
f′Fx(F) it is opposite k sections of beginning amounts of deflection caused by the concentrated force of section at x, For amount of deflection at x caused by kth section beginning corner.
Step 3, variation rigidity simply supported beam any position degree of disturbing under single form force effect；
Such as Fig. 8, when variation rigidity simply supported beam being asked to deform, simply supported beam can be converted to cantilever beam, a bearing of simply supported beam turns
Fixing end is turned to, another bearing is replaced with restraining force, and there are certain rotational angle thetas with former simply supported beam position for cantilever beamA, value is by beam
At restraining force end, amount of deflection is that zero condition is calculated.
Under unit concentrfated load effect, y indicates that concentrated force acts on distance of the simply supported beam position away from its left end, and l indicates letter
Distance between two fulcrums of strutbeam, if array z [0~H] is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam z [i-1]
Rigidity corresponding with z [i] sections is EI [i], then：
Then under unit concentrfated load effect, the amount of deflection w at the x of simply supported beam any positionF(x, y) is obtained by following formula：
Under unit concentrfated load effect, the corner γ at the x of simply supported beam any positionF(x, y) is obtained by following formula：
Under unit Uniform Loads, l indicates that the distance between two fulcrums of simply supported beam, L indicate simply supported beam overall length, array z
[0~M] is each section of rigidity initial position of variation rigidity cantilever beam, and cantilever beam z [i-1] rigidity corresponding with z [i] sections is EI [i], then：
Then under unit Uniform Loads, the amount of deflection w at the x of simply supported beam any positionq(x, L, l) is obtained by following formula：
Under unit Uniform Loads, the corner γ at the x of simply supported beam any positionq(x, L, l) is obtained by following formula：
Step 4, vehicle frame any position amount of deflection under single form force effect；
Using vehicle frame as research object, stressing conditions are applied to the power of vehicle frame as N as shown in figure 9, setting cargo fulcrumi(i=
1,2 ..., M), M counts for cargo branch, and each cargo fulcrum is Y away from axis vehicle front end distancei(i=1,2 ..., M).If array s [0
~P] it is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam s [i-1] rigidity corresponding with s [i] sections is GD [i]
It is intended to find out the power that each cargo fulcrum is applied to vehicle frame, is apparent from the problem and belongs to statically indeterminate problem, because large cargo is cut
Face the moment of inertia is far beyond vehicle frame the moment of inertia, it is therefore assumed that cargo is rigid body, the change of consideration vehicle frame main longitudinal grider with cargo support
The force structure of trailer main longitudinal grider under the more point actions of cargo can be then reduced to using cargo fulcrum as bearing by the influence of shape,
Using trailer main longitudinal grider as beam, using crossbeam to the active force and vehicle frame its own gravity of main longitudinal grider as the stress sketch of load, such as Figure 10
It is shown.
The superfluous constraint of cargo fulcrum at the 2nd to M-1 is released, so that statically indeterminate beam is become statically determinate beam, as shown in figure 11.
With fF(x, y) indicates to act on the unit concentrated force of vehicle frame y location that (cargo fulcrum is applied to the power N of vehicle frameiAnd
Each hydraulic axis support force Fj) load superposition can be used for vehicle frame any position degree of disturbing in caused amount of deflection at the x of vehicle frame position
Method is calculated, that is, calculates separately each power amount of deflection caused by same position, these amounts of deflection and be the final amount of deflection in the position.
The even distributed force formed to the weight of vehicle frame itself acts on vehicle frame situation, can be analyzed to Figure 12, Figure 13 and Figure 14 institute
Show that three kinds of stressing conditions, vehicle frame any position amount of deflection should be three kinds of stressing conditions the sum of caused amount of deflection in the position:
That is unit even distributed force amount of deflection caused by x at an arbitrary position：
fQ(x)=fQ1(x)+fQ2(x)+fQ3(x)
Wherein, fQ(x) amount of deflection caused by indicating unit even distributed force at x, fQ1(x)、fQ2(x)、fQ3(x) figure is indicated respectively
12, amount of deflection caused by Figure 13 and Figure 14 units even distributed force is at x (notices that Figure 14 units are evenly distributed with force direction and Figure 12, Figure 13 unit
Uniformly distributed force direction is opposite),
Step 5, each cargo fulcrum is asked to be applied to the power N of vehicle framei
Be zero for deformation compatibility condition with the amount of deflection at superfluous constraint, in conjunction be with joint efforts zero with to arbitrary point torque and be zero
Equilibrium condition then has：
In formula,It indicates
The power N of vehicle frame is applied to by cargo fulcrumi(i=1,2 ..., M), each hydraulic axis support force Fj(j=1,2 ..., N) and vehicle frame
The even distributed force that itself weight is formed at the superfluous constraint of statically indeterminate beam caused by amount of deflection be zero, M-2 can be constructed with this
A equation, to which equation group (2) can solve the power N that each cargo fulcrum is applied to vehicle framei(i=1,2 ..., M).(pay attention to：If meter
Certain the cargo fulcrum calculated is applied to the power N of vehicle frameiIt is negative, then it represents that the holder does not have supporting goods, should cast out this branch
Point recalculates)
Step 6, vehicle frame any position shearing force, moment of flexure and amount of deflection are asked；
If vehicle frame is Q (x) by shearing for institute at x away from axis vehicle front end distance, then have：
Wherein：
If vehicle frame is M (x) for the interfaces x institute bending moment away from axis vehicle front end distance, then have：
If vehicle frame is that amount of deflection is ω at x away from axis vehicle front end distancex, then have：
Step 7, vehicle frame maximum support and reinforcing beam setting；
By calculating, trailer main longitudinal grider usually will produce deformation up and down, and deflection, which crosses senior general, makes trailer hydraulic pressure
Suspension regulating power reduce, influence trailer longitudinally through ability, within the allowable range, to the convex situation of longeron, increasing can be taken
The method of holder or two side stands salient point movement upward is added to reduce convex deformation；To the recessed situation of longeron, it can take plus pre- arch subtracts
Few concave deformation, method and steps are as follows：
1, two outermost end position of the fulcrum are set.The most outer segment fulcrum of cargo two should ensure that there is certain distance apart from cargo both ends,
If the distance is ldlAnd ldr；
2, each fulcrum stress is calculated.In the fulcrum of both ends, if certain fulcrum stress is negative and there are certain fulcrum previous steps to be
Fulcrum is added, then cancels the added fulcrum of previous step, fulcrum stress is recalculated, in next step in record revocation position；
3, vehicle frame amount of deflection everywhere is calculated.If somewhere amount of deflection is just and to be more than maximum allowable positive amount of deflection Δ laIf do not exist herein
Position queue is cancelled, and the position is more than away from closest pivot distance between fulcrum allows distance la, then fulcrum is added at this,
2 are gone to, otherwise, in next step；
4, vehicle frame amount of deflection everywhere is calculated, is positive value in each deflection value and is more than maximum allowable positive amount of deflection Δ laIf the position
Closest two pivot distances are set more than permission distance l between fulcruma, then closest two fulcrum is towards position movement Δ l, when mobile
Two pivot distances should be not less than la；
5, vehicle frame amount of deflection everywhere is calculated, if somewhere amount of deflection is negative and more than the maximum allowable negative amount of deflection Δ l of regulationb, then at this
Place adds pre- arch, predomed height to be equal to amount of deflection and maximum allowable negative amount of deflection Δ l at this in sectionbThe absolute value of difference.
When the longitudinal strength deficiency of trailer, longitudinally reinforced girder steel can be installed additional on trailer to improve longeron intensity, installed additional
Reinforcing beam and trailer main longitudinal grider can only be stacked together, two beams have respective neutral axis respectively in bending deformation process.
Under normal circumstances, since trailer beam is very long, deforming latter two beam still can be close together, thus reinforcing beam and girder can be seen
Integrally, it is assumed that the moment of inertia increases Δ I after trailer girder adds reinforcing beam, judges whether that the method for adding reinforcing beam is as follows：
Vehicle frame moment of flexure everywhere is calculated, since longeron one end, is added successively in the longeron section setting that moment of flexure is more than moment of flexure allowable
Brutal：If certain section of moment of flexure is more than to allow moment of flexure, reinforcing beam is set in the section, trailer girder adds increased inertia after reinforcing beam
Square is：((this section of maximal bending moment-moment of flexure allowable)/moment of flexure allowable) * safety factor * trailer girder the moment of inertia * trailer girders are allowable
Stress and reinforcing beam allowable stress ratio.
Reinforcing beam both ends should at least exceed this section of both ends certain distance Δ lc。
Embodiment 1
The trailer number of axle is 20, wheelbase 1500mm, there is 4 supporting beams, cargo mass G=510t, each fulcrum of cargo away from
The distance of center of gravity of goods is respectively 5000mm, 2000mm, -2000mm, and -5000mm, (preceding just to bear afterwards), center of gravity of goods and trailer are vertical
In to just, vehicle frame weight is Gf=80t.Known cargo uses supported at three point, and preceding 2 latter points, the number of axle is respectively I=13, J
=7, trailer main longitudinal grider the moment of inertia is I=6.58 × 109mm4。
Seek each hydraulic axis support force Fj；
F is known by formula 11=... ..=F13, F14=... .=F20
So
I*F1+J*F20=G+Gf
Obtain F1=... .=F13=F14=...=F20=289.1KN
Each cargo fulcrum is asked to be applied to the power N of vehicle framei；
As m=2,
As m=3
So by
It obtains
N1=5078.24KN, N2=-2579.24KN, N3=-2579.24KN, N4=5078.24KN
It is negative force by C2, C3 fulcrum power are calculated above, shows that actually two fulcrum does not stress.It takes
N2、N3=0
Then N1,
Ask vehicle frame relevant position shearing force, moment of flexure and any position amount of deflection；
By formula (19), (20), acquires vehicle frame key position shearing and moment of flexure is as shown in table 1, it is left which calculates vehicle frame
Half portion key position shears and moment of flexure, since vehicle frame is along central symmetry stress, so right side shearing and moment of flexure size and left half
Portion's corresponding position shearing and moment of flexure size are identical.
As can be seen from the data in the table, moment M is in C1And C4Place is maximum, and maximal bending moment is
8305.9kNm, this example trailer longitudinal beam allows moment of flexure to be 6000kNm, therefore moment of flexure is more than to allow moment of flexure section (P7
Previous rice is to C15Latter rice section) reinforcing beam should be added, increased the moment of inertia is after adding reinforcing beam：
Wherein, safety factor takes 1.5, and trailer girder allowable stress is 1.5 with reinforcing beam allowable stress ratio.
Vehicle rail stress is recalculated, by formula (21), it is as shown in table 1 to acquire vehicle frame key position amount of deflection, right side
Amount of deflection size is identical as left side corresponding position amount of deflection.Data are shown in table, and trailer maximum distortion is 233mm in the middle part of trailer,
Since trailer deforms more than prescribed limit, it is contemplated that longitudinally add pre- arch 233mm, while reinforcing beam before load cargo for trailer longitudinal beam
Also make pre- arch, pre- arch value is 233-153=80mm, can by between reinforcing beam and trailer gasket realize.
1 vehicle frame key position amount of deflection of table
Those of ordinary skill in the art will understand that the embodiments described herein, which is to help reader, understands this hair
Bright implementation, it should be understood that protection scope of the present invention is not limited to such specific embodiments and embodiments.Ability
The those of ordinary skill in domain can make its various for not departing from essence of the invention according to the technical disclosures disclosed by the invention
Its various specific variations and combinations, these variations and combinations are still within the scope of the present invention.
Claims (2)
1. a kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety analytical method, which is characterized in that include the following steps：
Step 1, large cargo transports axis vehicle main beam stress static indeterminacy mechanical analysis；
With vehicle goods generally research object, if each hydraulic axis support force is respectively：Fj(j=1,2 ..., N),
If axis vehicle supported at three point marshalling using 2 points in preceding, number of axle I, a little rear, the number of axle J, I+J=N are then established such as
Lower equation group is to find out each hydraulic axis support force：
Goods weight is G in formula, and center of gravity of goods is L apart from vehicle frame front end distanceg, vehicle frame weight is Gf, vehicle frame length is L；Axis
Each wheelbase axis vehicle of vehicle front end distance is Xj(j=1,2 ..., N), N are the hydraulic pressure support number of axle；
If supported at three point using any preceding, number of axle I, 2 points in rear, number of axle J, calculation formula is identical；
Step 2, variation rigidity cantilever beam any position degree of disturbing under single form force effect；
If x is cantilever beam length, EI is cantilever beam bending stiffness；
Unit concentrfated load amount of deflection f caused by free end, rotational angle theta calculation formula are respectively：
Specific torque amount of deflection f caused by free end, rotational angle theta calculation formula are respectively：
It is respectively in free end amount of deflection f, rotational angle theta calculation formula caused by unit even distributed force：
For variation rigidity cantilever beam, if x indicates that, away from fixing end distance on cantilever beam, y indicates that concentrated force acts on cantilever beam position
Distance away from fixing end, if array z [0~H] is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam z [i-1] and z [i]
The corresponding rigidity of section is EI [i], then under unit concentrfated load effect, away from the rotational angle theta at fixing end distance x on cantilever beamFxIt can adopt
It is calculated with following formula：
If x is located at variation rigidity, cantilever beam kth section has if x≤y：
Wherein：
θ′FiIt is opposite this section of beginning corner in i-th section of cantilever beam end, has：θ′Fi=θ 'Fi(Mi)+θ′Fi(F), θ 'Fi(Mi) it is i-th
With respect to this section of end beginning corner caused by section cantilever beam tail end section moment of flexure,θ′Fi(F) it is
With respect to this section of end beginning corner caused by i-th section of cantilever beam tail end section concentrated force,
θ′FxFor opposite kth section cantilever beam beginning corner at x,
Have
If x>Y has：θFx=θFy；
Under unit concentrfated load effect, away from the amount of deflection f at fixing end distance x on cantilever beamFxFollowing formula calculating can be used：
If x is located at variation rigidity, cantilever beam kth section has if x≤y：
Wherein：
fF′iIt is opposite this section of beginning amount of deflection in i-th section of cantilever beam end, has：
f′Fi(Mi) it is with respect to this section of end beginning amount of deflection caused by i-th section of cantilever beam tail end section moment of flexure,f′Fi(F) it is that end caused by i-th section of cantilever beam tail end section concentrated force relatively should
Section beginning amount of deflection, For i sections of amounts of deflection caused by i-th section of beginning corner；
f′FxFor opposite kth section cantilever beam beginning amount of deflection at x, have
f′Fx(Mx) it is opposite k sections of beginning amounts of deflection caused by section turn moment at x,f′Fx(F) it is
Opposite k sections of beginning amounts of deflection caused by the concentrated force of section at x, For
Amount of deflection at x caused by kth section beginning corner；
If x>Y has：fFx=fFy+θFy(x-y)；
Variation rigidity cantilever beam is by Uniform Loads, cantilever beam length l, then under unit Uniform Loads, on cantilever beam
Away from the rotational angle theta at fixing end distance xqx, obtained using following formula：
Wherein, θ 'qiIt is opposite this section of beginning corner in i-th section of cantilever beam end, has：θ′qi=θ 'qi(q)+θ′qi(Mi)+θ′qi(F),
θ′qi(q) it is with respect to this section of end beginning corner caused by i-th section of cantilever beam end even distributed force,θ′qi
(Mi) with respect to this section of end beginning corner caused by i-th section of cantilever beam tail end section moment of flexure,
θ′qi(F) with respect to this section of end beginning corner caused by i-th section of cantilever beam tail end section concentrated force,
Wherein, θq′xFor opposite k sections of beginning corners at x, have：θ′qx=θ 'qx(q)+θ′qx(Mi)+θ′qx(F), θ 'qx(q) it is phase at x
To opposite k sections of beginning corners caused by k sections of beginning even distributed forces,θ′qx(Mi) it is that section is curved at x
Opposite k sections of beginning corners at x caused by square,θ′qx(F) it is that section concentrated force draws at x
Opposite k sections of beginning corners at the x risen,
Under unit Uniform Loads, away from the amount of deflection f at fixing end distance x on cantilever beamqxIt is obtained using following formula：
Wherein, f 'qiIt is opposite this section of beginning amount of deflection in i-th section of cantilever beam end, has：
f′qi(q) it is with respect to this section of end beginning amount of deflection caused by i-th section of cantilever beam end even distributed force,
f′qi(Mi) with respect to this section of end beginning caused by i-th section of cantilever beam tail end section moment of flexure scratches
Degree,fq′i(F) end phase caused by i-th section of cantilever beam tail end section concentrated force
To this section of beginning amount of deflection, Cause for i-th section of beginning corner
I section amounts of deflection；
f′qxFor opposite kth section cantilever beam beginning amount of deflection at x, have f′qx(q) it is opposite k sections of beginning amounts of deflection caused by even distributed force at x,
f′Fx(Mx) it is opposite k sections of beginning amounts of deflection caused by section turn moment at x,f′Fx(F)
For opposite k sections of beginning amounts of deflection caused by section concentrated force at x,
For amount of deflection at x caused by kth section beginning corner；
Step 3, variation rigidity simply supported beam any position degree of disturbing under single form force effect；
When variation rigidity simply supported beam being asked to deform, simply supported beam is converted to cantilever beam, a bearing of simply supported beam is converted into fixing end, separately
One bearing is replaced with restraining force, and there are certain rotational angle thetas with former simply supported beam position for cantilever beamA, value scratched by beam at restraining force end
Degree is that zero condition is calculated；
Under unit concentrfated load effect, y indicates that concentrated force acts on distance of the simply supported beam position away from its left end, and l indicates simply supported beam
Distance between two fulcrums, if array z [0~H] is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam z [i-1] and z
It is EI [i] that [i] section, which corresponds to rigidity, then：
Then under unit concentrfated load effect, the amount of deflection w at the x of simply supported beam any positionF(x, y) is obtained by following formula：
Under unit concentrfated load effect, the corner γ at the x of simply supported beam any positionF(x, y) is obtained by following formula：
Under unit Uniform Loads, the distance between l expression two fulcrums of simply supported beam, L expression simply supported beam overall lengths, array z [0~
M] it is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam z [i-1] rigidity corresponding with z [i] sections is EI [i], then：
Then under unit Uniform Loads, the amount of deflection w at the x of simply supported beam any positionq(x, L, l) is obtained by following formula：
Under unit Uniform Loads, the corner γ at the x of simply supported beam any positionq(x, L, l) is obtained by following formula：
Step 4, vehicle frame any position amount of deflection under single form force effect；
Using vehicle frame as research object, if the power that cargo fulcrum is applied to vehicle frame is
Ni(i=1,2 ..., M), M count for cargo branch, and each cargo fulcrum is Y away from axis vehicle front end distancei(i=1,2 ..., M)；
If array s [0~P] is each section of rigidity initial position of variation rigidity cantilever beam, cantilever beam s [i-1] rigidity corresponding with s [i] sections
For GD [i]；
The power that each cargo fulcrum is applied to vehicle frame is found out, because large cargo cross sectional moment of inertia is far beyond vehicle frame the moment of inertia, therefore
It is assumed that cargo is rigid body with cargo support, the influence of the deformation of consideration vehicle frame main longitudinal grider then will be under the more point actions of cargo
The force structure of trailer main longitudinal grider is reduced to using cargo fulcrum as bearing, using trailer main longitudinal grider as beam, with crossbeam to main longitudinal grider
Active force and the stress sketch that vehicle frame its own gravity is load；
The superfluous constraint of cargo fulcrum at the 2nd to M-1 is released, statically indeterminate beam is made to become statically determinate beam；
With fFAmount of deflection caused by (x, y) indicates to act on the unit concentrated force of vehicle frame y location at the x of vehicle frame position, appoints vehicle frame
Meaning position degree of disturbing, can be used Load superposition and is calculated, that is, calculate separately each power amount of deflection caused by same position, these are scratched
Degree and be the final amount of deflection in the position；
The even distributed force formed to the weight of vehicle frame itself acts on vehicle frame situation, decomposes three kinds of stressing conditions, vehicle frame any position
The sum of amount of deflection caused by amount of deflection should be three kinds of stressing conditions in the position:
That is unit even distributed force amount of deflection caused by x at an arbitrary position：
fQ(x)=fQ1(x)+fQ2(x)+fQ3(x)
Wherein, fQ(x) amount of deflection caused by indicating unit even distributed force at x, fQ1(x)、fQ2(x)、fQ3(x) indicate that unit is uniformly distributed respectively
Power caused amount of deflection at x；
Step 5, each cargo fulcrum is asked to be applied to the power N of vehicle framei；
Be zero for deformation compatibility condition with the amount of deflection at superfluous constraint, in conjunction be with joint efforts zero with to arbitrary point torque and be zero balancing
Condition then has：
In formula,It indicates by goods
Object fulcrum is applied to the power N of vehicle framei(i=1,2 ..., M), each hydraulic axis support force Fj(j=1,2 ..., N) and vehicle frame itself
The even distributed force that is formed of weight at the superfluous constraint of statically indeterminate beam caused by amount of deflection be zero, can construct M-2 just with this
Journey, to which equation group (2) can solve the power N that each cargo fulcrum is applied to vehicle framei(i=1,2 ..., M)；
Step 6, vehicle frame any position shearing force, moment of flexure and amount of deflection are asked；
If vehicle frame is Q (x) by shearing for institute at x away from axis vehicle front end distance, then have：
Wherein：
If vehicle frame is M (x) for the interfaces x institute bending moment away from axis vehicle front end distance, then have：
If vehicle frame is that amount of deflection is ω at x away from axis vehicle front end distancex, then have：
Step 7, vehicle frame maximum support and reinforcing beam setting；
To the convex situation of longeron, the method for increasing the salient point movement upward of holder or two side stands is taken to reduce convex deformation；To vertical
The recessed situation of beam, takes plus pre- arch reduces concave deformation, and steps are as follows：
71, two outermost end position of the fulcrum are set, the most outer segment fulcrum of cargo two should ensure that there is certain distance apart from cargo both ends, if should
Distance is ldlAnd ldr；
72, each fulcrum stress is calculated, in the fulcrum of both ends, if certain fulcrum stress is negative and is addition there are certain fulcrum previous step
Fulcrum, then cancel the added fulcrum of previous step, and fulcrum stress is recalculated, in next step in record revocation position；
73, vehicle frame amount of deflection everywhere is calculated, if somewhere amount of deflection is just and to be more than maximum allowable positive amount of deflection Δ laIf do not removing herein
Position queue is sold, and the position is more than away from closest pivot distance between fulcrum allows distance la, then fulcrum is added at this, is gone to
72, otherwise, in next step；
74, vehicle frame amount of deflection everywhere is calculated, is positive value in each deflection value and is more than maximum allowable positive amount of deflection Δ laIf the position is most
Neighbouring two pivot distances allow distance l between being more than fulcruma, then closest two fulcrum is towards position movement Δ l, two when mobile
Point distance should be not less than la；
75, vehicle frame amount of deflection everywhere is calculated, if somewhere amount of deflection is negative and more than the maximum allowable negative amount of deflection Δ l of regulationb, then in the place
In section plus pre- arch, predomed height is equal to amount of deflection and maximum allowable negative amount of deflection Δ l at thisbThe absolute value of difference.
2. a kind of highway awkward and lengthy cargo transport trailer main longitudinal grider safety analytical method according to claim 1, feature
It is：When the longitudinal strength deficiency of trailer, longitudinally reinforced girder steel is installed additional on trailer to improve longeron intensity, the reinforcement of installation
Beam and trailer main longitudinal grider can only be stacked together, and two beams have respective neutral axis respectively in bending deformation process；General feelings
Under condition, since trailer beam is very long, deforming latter two beam still can be close together, thus reinforcing beam and girder can be regarded as one
Body, it is assumed that the moment of inertia increases Δ I after trailer girder adds reinforcing beam, judges whether that the method for adding reinforcing beam is as follows：
Vehicle frame moment of flexure everywhere is calculated, since longeron one end, reinforcing beam is set in the longeron section that moment of flexure is more than moment of flexure allowable successively：
If certain section of moment of flexure is more than to allow moment of flexure, reinforcing beam is set in the section, increased the moment of inertia is after trailer girder addition reinforcing beam：
((this section of maximal bending moment-moment of flexure allowable)/moment of flexure allowable) * safety factor * trailer girder the moment of inertia * trailer girder allowable stresses with
Reinforcing beam allowable stress ratio；Reinforcing beam both ends should at least exceed this section of both ends certain distance Δ lc。
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| CN112199774A (en) * | 2020-09-27 | 2021-01-08 | 一汽解放青岛汽车有限公司 | Method for calculating bending deflection of commercial vehicle frame in whole vehicle state |
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| CN111967106A (en) * | 2020-08-19 | 2020-11-20 | 深圳市康冠科技股份有限公司 | Method for determining length size of irregular object base |
| CN111967106B (en) * | 2020-08-19 | 2024-01-09 | 深圳市康冠科技股份有限公司 | Method for determining length dimension of irregular object base |
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| CN112199774B (en) * | 2020-09-27 | 2022-11-25 | 一汽解放青岛汽车有限公司 | Method for calculating bending deflection of commercial vehicle frame in whole vehicle state |
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