CN111428319A - Circular cross section equal strength supporting beam with evenly distributed load at intervals - Google Patents
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Abstract
A supporting beam with uniform strength and circular cross section and evenly distributed loads at intervals belongs to the technical field of mechanical structure design and manufacturing, and solves the technical problems that beam materials cannot be fully utilized, the structural size is too large and the like due to the fact that internal loads of the beam are unevenly distributed. The structural dimension of the circular cross section equal-strength supporting beam subjected to the loads evenly distributed at intervals changes along with the size of the internal load, so that the maximum stress on each cross section of the beam is equal, and the technical effects of equal strength of each cross section, material saving, simplified structure, weight reduction and the like are achieved.
Description
Technical Field
The invention relates to a circular cross section equal strength supporting beam evenly loaded at intervals, belonging to the technical field of mechanical structure design and manufacture.
Background
In the production practice, the technical problem of girders bearing evenly-spaced loads is often encountered, for example, aluminum foil in an aluminum foil production plant needs to be supported by the girders in the processes of heat treatment and the like. The aluminum foil is produced in rolls, the aluminum foil is wound around a hollow core, an aluminum roll is about 300kg to 2000kg, and a plurality of aluminum foils are supported through the core by a girder in a heat treatment furnaceThe aluminum foil rolls are arranged at certain intervals, and as shown in figure 1, the aluminum foil roll support is a schematic view and has a length of L2The girder through the core supports a number of aluminium foil coils, equivalent to bearing a load evenly spaced, the spacing between the aluminium foil coils being a, a mechanical model of such engineering problems can be represented by fig. 2. fig. 2 is a mechanical model suitable for any similar engineering problem bearing a load beam evenly spaced, fig. 2, L1The length distributed by the uniformly distributed load, q the concentration or linear density (unit is N/m or kN/m) of the uniformly distributed load, x the distance from any section of the beam AB to the left end A of the beam AB, and V the counter force of the support.
Because the external load is very large and the action mode of the external load is also special, the problems of strength and rigidity of the supporting beam become engineering technical problems to be solved urgently. The traditional design mostly determines the structural size of the beam by the maximum stress of the most dangerous section of the limited beam, and the size of the whole beam is the same as that of the most dangerous section, so that materials do not fully play a role in the position with small stress, waste is caused, the weight of the beam is increased, the cost is increased, and the operation and the use are difficult. A more scientific and reasonable idea is to determine reasonable shapes and sizes of the components at the most economical cost on the premise of ensuring safety and reliability. For the circular cross section equal-strength supporting beam bearing the loads uniformly distributed at intervals, if a beam with a special structure size can be invented, all cross sections of the beam are under the same strength, the maximum stress positions of all cross sections are equal, and the technical effects of safety, reliability, science and economy can be achieved. This can be done according to the relevant expertise, and a specific analysis will be made below. In an era of increasingly advanced numerical control machining technology, such a structure is also easy to machine and manufacture.
Disclosure of Invention
The technical scheme adopted by the invention for solving the technical problems is as follows: the invention relates to a circular cross section equal strength supporting beam evenly distributed with load at intervals, which is characterized in that: the beam bears evenly spaced loads, i.e. evenly distributed loads per unit length along the length of the beam at regular intervals, the cross-section of the beam being solidThe radius of the circle is changed along with the length direction x of the beam, namely the radius of the circle at the position x away from the left end of the beam is r (x); (1) when the cross section of the beam is a solid circle, the radius of the cross section of the beam in the interval section (section without external load) is
Wherein
k is the number of the beam spacer (no external load), k is odd number, a is the length of the beam spacer (no external load), L1The length of each section of the uniformly distributed load, Q is the total load (i.e. the resultant force of all distributed loads) borne by the beam, n is the number of sections of the uniformly distributed load borne by the beam, and [ sigma ]]Allowable stress for tensile and compressive of the beam material; radius of circular cross section of beam in section bearing uniformly distributed load
Wherein
k is the number of the sections of the beam bearing the evenly distributed load sections, k is an even number, a is the length of the beam spacing section (no external load section), L1The length of each section of the uniformly distributed load, Q is the total load (i.e. the resultant force of all distributed loads) borne by the beam, n is the number of sections of the uniformly distributed load borne by the beam, and [ sigma ]]Allowable stress for tensile and compressive of the beam material; (2) when the cross section of the beam is a circular ring, the outer radius r (x) of the circular ring meets the following structural size constraint condition:
wherein t is the width of the ring (i.e., the difference between the outer radius and the inner radius of the ring), [ sigma ]]Allowable tension and compression stress, M, for beam materialk(x) For the bending moment of the beam at the x-section, Mk(x) The following two cases are distinguished: (i) when M isk(x) When the beam spacing segment (without external load segment) bears bending moment and the segment number mark k is odd number,
wherein
a is the length of the beam spacing section (no external load section), L1The length of each section of the uniformly distributed load, Q is the total load (namely the resultant force of all the uniformly distributed loads) borne by the beam, and n is the number of sections of the uniformly distributed load borne by the beam; (ii) when M isk(x) When the beam bears the bending moment borne by the evenly distributed load sections and the section number k is an even number,
the other symbols have the same meanings as above. The beam bears loads which are evenly distributed at intervals
The meanings of the symbols are as defined above. When the cross section of the beam is a solid circular cross section, the minimum radius of the beam is
And r isminT is more than or equal to t; when the cross-section of the beam is a hollow circular cross-section (circular ring), the minimum radius of the beam is
And c isminT is more than or equal to t; wherein [ tau ] is]The shear allowable stress of the beam material is defined, and the rest symbols are as before.
The invention has the beneficial effects that: the beam has equal strength everywhere, saves materials, simplifies the structure and reduces weight.
Drawings
FIG. 1 schematic view of a roll of aluminum foil support
FIG. 2 is a diagram of a mechanical model of spaced load beams
Figure 3 equal strength supporting beam under second working condition of embodiment
FIG. 4 shows an embodiment of an equal-strength supporting beam under three working conditions
In fig. 1: 1 is a support beam, 2 is a roll of aluminum foil.
L in FIG. 21Length distributed for uniform load distributionQ is the concentration or linear density (unit is N/m or kN/m) of the uniform load, L2The total length of the beam, x is the distance from any section of the beam AB to the left end A of the beam AB, V is the counter force of the support, Q is the total load born by the beam (namely the resultant force of the uniformly distributed loads of all sections on the beam), and a is the interval between all the aluminum foil rolls.
Detailed Description
To achieve equal strength, the external load characteristics, the internal force (moment) characteristics and rules of the beam and the change trend and rules of each parameter are analyzed and researched, and corresponding technical measures are taken according to research results. The technical scheme and the technical measures of the invention are also quite special due to the special external load of the beam.
As shown in FIG. 2, AB represents a supporting beam with a circular cross section and evenly distributed loads, and the total length of the beam is L2The total force of n sections of evenly distributed loads, namely the total load born by the beam, is Q N, the concentration degree of each section of evenly distributed load, namely the linear density is q N/m, and the distribution length of each section of evenly distributed load is L1The interval between every two evenly distributed loads is a; according to the relevant expert knowledge, the internal moment on a cross section at x from the left end of the beam (see fig. 2) can be found as follows:
(i) bending moment of interval section (section without external load)
When M isk(x) When the beam spacing segment (without external load segment) bears bending moment and the segment number mark k is odd number,
wherein k is a segment number index, and k is an odd number,
The other symbols have the same meanings as above. (ii) Bending moment of beam in section bearing uniformly distributed load
When M isk(x) When the beam bears the bending moment borne by the evenly distributed load sections and the section number k is an even number,
wherein k is the number of stages and is an even number,
The other symbols have the same meanings as above.
Through analytical research, the following characteristics can be obtained:
(1) the internal bending moment of the load beam uniformly distributed at intervals has symmetry, namely M1(x)＝M2n+1(L2-x)、M2(x)＝M2n(L2-x)、M3(x)＝M2n-1(L2-x)、M4(x)＝M2n-2(L2-x)...Mk(x)＝M2n-k+2(L2-x)。
(2) When bearing even load of even number section (k is odd number), the maximum bending moment is at the middle point of the middle interval section of the whole beam and is constant
In this case, k is an odd number, and k is n + 1.
(3) When bearing odd-number evenly-distributed loads (k is even number), the maximum bending moment is at the middle point of the aluminum coil in the middle of the whole beam
If it is
Then
K is an even number, and k is n + 1; the above-mentioned
The maximum bending moment of the whole beam under the action of completely uniformly distributed loads is obtained.
(4) In conclusion, when the load is uniformly distributed on the n sections, the interval section has n +1 sections and 2n +1 sections in total, and the number of bending moments is 2n + 1; when n is odd number, the maximum bending moment is even number bending moment Mk＝Mn+1(i.e., k is n +1), which occurs at the midpoint of the evenly-distributed load segment in the middle, i.e., the midpoint of the (n +1) th segment, i.e., the middle point
Where, i.e. the mid-point of the full beam, the maximum bending moment is
Incidentally, if L1＝L2This means that a is 0 and n is 1, and the above formula is expressed in this case
This is the maximum bending moment known for a full beam when fully loaded evenly (without spacing). When n is even number, the maximum bending moment is odd number bending moment Mk＝Mn+1(i.e., k ═ n +1), occurs in the (n +1) th compartment and is constant throughout the (n +1) th compartment, i.e., from
To
Upper constant, maximum bending moment is
In addition, for odd numbered (spaced) bending moments
This is a linear equation, and it is clear that k<n +1, MkRises as x increases; k is a radical of>n +1, MkDecreases as x increases; when k is n +1, MkIs constant, independent of x, is
Completely coincide with the above.
For even number, i.e. uniformly distributing load section bending moment
This is a parabolic equation, let
To obtain
To make x0Fall on
In the interior, it is required to have
n<k<n +2, where n and k are positive integers, so that k is n +1, and when k is n +1,
i.e. at the midpoint of the entire beam. Maximum value of
If the relation of k ═ n +1 is not satisfied, MkIs not at the maximum value of
And (4) the following steps. Since k is an even number, n is an odd number. Also in full agreement with the foregoing.
Similar conclusions are also drawn for the radius of the beam, i.e. the beam height, namely:
radius for odd-numbered segments (interval segments, i.e. segments without external load)
k<n +1, rk(x) Rises as x increases; k is a radical of>n +1, rk(x) Decreases as x increases; when k is n +1, rk(x) Is a constant, independent of x,is composed of
For even numbers, i.e. radii of evenly-distributed load segments
This is in conjunction with Mk(x) Is identical, let x be0Fall on
Within, n is required<k<n +2, where n and k are positive integers, so that k is n +1, and when k is n +1,
i.e. at the midpoint of the entire beam. If the relation of k ═ n +1 is not satisfied, rkIs not at the maximum value of
And (4) the following steps. Since k is an even number, n is an odd number. When in use
When r iskIs at a maximum of
Obviously, if the size of the beam is determined by the maximum bending moment of the whole beam, the beam is necessarily large (see the following embodiment), which is neither economical nor safe, because the structure is heavy, which wastes manpower, material resources and financial resources, and the operation and use are inconvenient. The technical idea of the invention is to adapt the structural size of the beam to the special change of the internal moment so as to achieve the technical effect of equal strength of the beam everywhere.
In the first embodiment, if the number of evenly distributed load sections n on the solid circular-section beam is 5, the maximum sequence number of the odd-numbered bending moment is M2n+1I.e. M11The maximum sequence number of even-number bending moment is M2nI.e. M10Its 5-segment uniform load total load Q is 73.5kN, so that its support reaction force V is Q/2 is 36.75kN, and its beam length is L2Each segment is uniformly distributed with intervals between the loads of 3m
L1Equal load of 564mm
According to the formula (1) (odd number section, namely bending moment of the interval section) and the formula (2) (even number section, namely bending moment of the evenly distributed load section) and the technical scheme of the invention, the bending moment M of each section can be obtainedk(x) (kNm) and radial dimension r of the beamk(x) (m) is as follows:
x is more than or equal to 0 and less than or equal to a in the 1 st section, namely x is more than or equal to 0 and less than or equal to 0.03m in the section, k is 1:
M1(x)＝Vx＝Qx/2＝36.75x；
it is possible to verify that,
i.e. the maximum stress on each cross-section is constant over the first section [ sigma ]]Wherein W is1Is the bending section coefficient of the first segment circular cross section,
x is more than or equal to a and less than or equal to a + L in the 2 nd section1That is, x is 0.03m ≦ 0.594m, k is 2:
it is possible to verify that,
i.e. the maximum stress in each cross-section is constant over the entire second section [ sigma ]]Wherein W is2The bending section coefficient of the second section of circular cross section,
the starting point is denoted by s and the end point by u, and it can be verified that when x is a, M is1u＝M2s。
Section 3a + L1≤x≤2a+L1I.e. x is more than or equal to 0.594m and less than or equal to 0.624m, k is 3:
it can be verified that x is 2a + L1When M is in contact with2u＝M3s。
It is possible to verify that,
i.e. the maximum stress in each cross-section is constant over the third section [ sigma ]]Wherein W is3Is the bending section coefficient of the circular cross section of the third section,
by the same token, it can be verified that the following sections all have
I.e. the maximum stress on each cross-section of the segments is constant [ sigma ]]Wherein
Is a k-th segment of a circleBending section modulus of the cross section. This is an advantage of equal strength.
4 th stage 2a + L1≤x≤2a+2L1I.e. x is more than or equal to 0.624m and less than or equal to 1.188m, k is 4:
x＝2a+L1when M is in contact with3u＝M4s(ii) a And, M4max＝M4u。
Segment 5, 2a + 2L1≤x≤3a+2L1Namely, x is more than or equal to 1.188m and less than or equal to 1.218m, k is 5:
it can be verified that x is 2a + 2L1When M is in contact with4u＝M5s。
6 th stage 3a + 2L1≤x≤3a+3L1Namely, x is more than or equal to 1.218m and less than or equal to 1.782m, k is 6:
it can be verified that x is 3a + 3L1When M is in contact with5u＝M6s＝M5max。
Order to
To obtain
Mmax＝M6max＝Q(0.9a+0.625L1)＝27.89325kNm。
7 th stage 3a + 3L1≤x≤4a+3L1Namely, x is more than or equal to 1.782m and less than or equal to 1.812m, k is 7:
9 th stage 4a + 4L1≤x≤5a+4L1Namely 2.376m is less than or equal to x is less than or equal to 2.406m, k is 9:
11 th stage 5a + 5L1≤x≤6a+5L1Namely, x is more than or equal to 2.97m and less than or equal to 3m, k is 11:
8 th stage 4a + 3L1≤x≤4a+4L1Namely 1.812m is less than or equal to x is less than or equal to 2.376m, k is 8:
10 th stage 5a + 4L1≤x≤5a+5L1Namely 2.406m is less than or equal to x is less than or equal to 2.97m, k is 10:
maximum bending moment M if using whole beammax＝M6maxWhen the beam height is controlled to 27.89325kNm, the whole beam has a constant radius
If take [ sigma ]]＝170000kN/m2B 40mm, the diameter of each cross section of the whole beam must be 2r 118.6725033mm, which is the diameter of the beam at the highest point of the beam:
taking the allowable shearing stress [ tau ]]Minimum radius at 90MPa
The beam height of the constant-strength beam is 2rk(x) I.e. beams having the above characteristics and rules, and the shape and size of which conform to the above characteristics and rules, must be the same everywhere, i.e. the maximum stress of each cross section is equal. The same applies to the second and third embodiments, and the second and third embodiments also give equal strength verification.
In the second embodiment, if the number of evenly distributed load sections n on the solid circular-section beam is 5, the maximum number of the odd-numbered bending moments is M2n+1I.e. M11The maximum sequence number of even-numbered bending moment isM2nI.e. M10Its 5-segment uniform load total load Q is 73.5kN, so that its support reaction force V is Q/2 is 36.75kN, and its beam length is L2Each segment is uniformly distributed with intervals between the loads of 3m
L1Equal to 360mm, load is evenly distributed
According to the formula (1) (odd number section, namely bending moment of the interval section) and the formula (2) (even number section, namely bending moment of the evenly distributed load section) and the technical scheme of the invention, the bending moment M of each section can be obtainedk(x) (kNm) and radius r of the beamk(x) (m) is as follows (2 r)k(x) I.e., beam height).
(in the first and second embodiments, the external loads are the same, the expression forms of the bending moment and the radius are the same, but the intervals of the uniformly distributed loads are different, so that the section lengths suitable for the bending moment are different, and the sizes of the bending moment and the radius are different)
X is more than or equal to 0 and less than or equal to a in the 1 st section, namely x is more than or equal to 0 and less than or equal to 0.2m in the section, k is 1:
M1(x)＝Vx＝Qx/2＝36.75x；
it is possible to verify that,
i.e. the maximum stress on each cross-section is constant over the first section [ sigma ]]Wherein W is1Is the bending section coefficient of the first segment circular cross section,
x is more than or equal to a and less than or equal to a + L in the 2 nd section1I.e. x is more than or equal to 0.2m and less than or equal to 0.56m, k is 2:
it is possible to verify that,
i.e. the maximum stress in each cross-section is constant over the entire second section [ sigma ]]Wherein W is2Is the bending section coefficient of the cross section of the second section of circle,
the starting point is denoted by s and the end point by u, and it can be verified that when x is a, M is1u＝M2s。
Section 3a + L1≤x≤2a+L1I.e. x is more than or equal to 0.56m and less than or equal to 0.76m, k is 3:
it can be verified that x is 2a + L1When M is in contact with2u＝M3s。
It is possible to verify that,
i.e. the maximum stress in each cross-section is constant over the third section [ sigma ]]Wherein W is3Is the bending section coefficient of the circular cross section of the third section,
by the same token, it can be verified that the following sections all have
I.e. the maximum stress on each cross-section of the segments is constant [ sigma ]]Wherein
The bending section coefficient of the k-th section circular cross section is shown. This is achievedIs the advantage of equal strength.
4 th stage 2a + L1≤x≤2a+2L1I.e. x is more than or equal to 0.76m and less than or equal to 1.12m, k is 4:
x＝2a+L1when M is in contact with3u＝M4s(ii) a And, M4max＝M4u。
Segment 5, 2a + 2L1≤x≤3a+2L1Namely, x is more than or equal to 1.12m and less than or equal to 1.32m, k is 5:
it can be verified that x is 2a + 2L1When M is in contact with4u＝M5s。
6 th stage 3a + 2L1≤x≤3a+3L1Namely, x is more than or equal to 1.32m and less than or equal to 1.68m, k is 6:
it can be verified that x is 3a + 3L1When M is in contact with5u＝M6s＝M5max。
Order to
To obtain
Mmax＝M6max＝Q(0.9a+0.625L1)＝29.7675kNm
7 th stage 3a + 3L1≤x≤4a+3L1Namely, x is more than or equal to 1.68m and less than or equal to 1.88m, k is 7:
9 th stage 4a + 4L1≤x≤5a+4L1Namely, x is more than or equal to 2.24m and less than or equal to 2.44m, k is 9:
11 th stage 5a + 5L1≤x≤6a+5L1Namely, x is more than or equal to 2.8m and less than or equal to 3m, k is 11:
8 th stage 4a + 3L1≤x≤4a+4L1Namely, x is more than or equal to 1.88m and less than or equal to 2.24m, k is 8:
10 th stage 5a + 4L1≤x≤5a+5L1Namely, x is more than or equal to 2.44m and less than or equal to 2.8m, k is 10:
maximum bending moment M if using whole beammax＝M6maxThe beam radius is controlled to 29.7675kNm, and the diameter of the whole beam is constant
If take [ sigma ]]＝170000kN/m2The diameter of each cross section of the whole beam must be 121.2731108mm at 2r, which is the height of the beam with equal strength at the highest position:
taking the allowable shearing stress [ tau ]]Minimum radius at 90MPa
And r is 14mm at the head end and the tail end of the beam, namely a section of 0-10 mm and a section of 2990-3000 mm.
The shape and size of the beam are determined by the above expressions and data, and the beam having the above characteristics is necessarily an equal strength beam for the working conditions given in this embodiment. Fig. 3 is a schematic view of the constant-strength beam according to the second embodiment.
In the third embodiment, all working condition parameters are the same as those of the second embodiment, a hollow circular-section beam is used, the number of sections n for uniformly distributing loads is 5, and the maximum sequence number of odd-numbered bending moments is M2n+1I.e. M11The maximum sequence number of even-number bending moment is M2nI.e. M10Its 5-segment uniform load total load Q is 73.5kN, so that its support reaction force V is Q/2 is 36.75kN, and its beam length is L2Each segment is uniformly distributed with intervals between the loads of 3m
L1Equal to 360mm, load is evenly distributed
According to the formula (1) (odd number section, namely bending moment of the interval section) and the formula (2) (even number section, namely bending moment of the evenly distributed load section) and the technical scheme and measures of the invention, the bending moment M of each section can be obtainedk(x) (kNm) as described in example two. The outer radius r (x) of the hollow circular cross section (circle) meets the following structural dimensional constraints:
wherein t is the width of the circumference of the ring, [ sigma ]]Allowable tension and compression stress (same as example two), M for beam materialk(x) The bending moment of the beam at the x-section is exactly the same as in the second embodiment. The above constraints will now be further explained to show the specific implementation. Transforming the above formula into
Let A be t and B be-1.5 t2，
D＝－0.25t4，
Make it
Then there are:
(1) if Δ>0，
(2) If Δ is 0 and p ≧ 0,
or
(3) If Δ<0，
Or
Or
Wherein
Radius of beam rk(x) With the above rules and characteristics, it is necessary to have beams of equal strength, i.e. the beam strength is equal everywhere, i.e. the maximum stress on each cross section of the beam is equal.
Can select reasonable and optimal beam radius according to the rule and ensure the minimum radius
And r isminT is more than or equal to t. Taking the operating condition of this embodiment as an example, take t of the ring as 10mm, and pull-press allowable [ σ [ ]]170MPa, shear allowable stress [ tau ]]＝90MPa。
According to the technical scheme of the invention, the bending moment and the beam height of each equal-strength beam are as follows (the bending moments of the second embodiment and the third embodiment and the suitable section lengths thereof are completely the same, but the cross sections are different, so that the beam radiuses of the two embodiments are different). To show dimensional values, the following expression is not given for beam radius as in example two, but specific data are listed every 10mm along the beam length, see tables 1-4. In order to verify the technical effect of the present invention, several sets of data are randomly extracted to be examined as follows.
X is more than or equal to 0 in the 1 st sectionA is less than or equal to a, namely x is less than or equal to 0 and less than or equal to 0.2m, k is 1: m1(x)＝Vx＝Qx/2＝36.75x；
Section 3a + L1≤x≤2a+L1I.e. x is more than or equal to 0.56m and less than or equal to 0.76m, k is 3:
segment 5, 2a + 2L1≤x≤3a+2L1Namely, x is more than or equal to 1.12m and less than or equal to 1.32m, k is 5:
minimum radius
The radius r of the beam can be 22mm at the head end and the tail end of the beam, namely at the section of 0-30 mm and the section of 2970-3000 mm>t, see data in parentheses in tables 1 and 4.
And (3) verification:
(1) data are taken at 100mm in section 1 (see table 1): the beam radius r (100) 33.033mm (accurate data 33.03290622mm), bending moment m (x) 3.675kNm, the aforementioned fixed t10 mm. Thus, the
Substituting t 0.01m, r (x) 0.03303290622m to obtain W2.16176 × 10m-5m3。
Maximum stress of
(2) Data were also taken at 2000mm x in paragraph 8 (see table 4): the radius r (x 2000) of the beam is 77.599mm (accurate data is 77.59901235mm), the bending moment m (x) is 26.46kNm,
substituting t as 0.01m, r (x as 2000) as 0.07759901235m to obtain W as 0.000155647m3。
Maximum stress of
The stress at other positions can be calculated in the same way, and the result is that the maximum stress sigma is 170MPa everywhere. This is the equal strength advantage achieved by the present invention.
7 th stage 3a + 3L1≤x≤4a+3L1Namely, x is more than or equal to 1.68m and less than or equal to 1.88m, k is 7:
9 th stage 4a + 4L1≤x≤5a+4L1Namely, x is more than or equal to 2.24m and less than or equal to 2.44m, k is 9:
11 th stage 5a + 5L1≤x≤6a+5L1Namely, x is more than or equal to 2.8m and less than or equal to 3m, k is 11:
x is more than or equal to a and less than or equal to a + L in the 2 nd section1I.e. x is more than or equal to 0.2m and less than or equal to 0.56m, k is 2:
4 th stage 2a + L1≤x≤2a+2L1I.e. x is more than or equal to 0.76m and less than or equal to 1.12m, k is 4:
6 th stage 3a + 2L1≤x≤3a+3L1Namely, x is more than or equal to 1.32m and less than or equal to 1.68m, k is 6:
8 th stage 4a + 3L1≤x≤4a+4L1Namely, x is more than or equal to 1.88m and less than or equal to 2.24m, k is 8:
10 th stage 5a + 4L1≤x≤5a+5L1Namely, x is more than or equal to 2.44m and less than or equal to 2.8m, k is 10:
specific dimensions and shapes of the constant-strength beam are determined by the data in tables 1-4. Fig. 4 shows an equal-strength beam obtained under the working conditions of the third embodiment and the technical scheme of the invention. The beam height of example three is greater than that of example two, but since the beam of example three is hollow, its cross-sectional area is smaller, as can be verified using the beam height data of tables 1-4.
TABLE 1 numerical values of radius and bending moment of 1 st, 3 rd and 5 th sections of beams
TABLE 2 values of radius and bending moment of 7 th, 9 th and 11 th sections of beam
TABLE 3 radius and moment values of beams at 2 nd, 4 th and 6 th sections
TABLE 4 radius and bending moment values of the beams at sections 8 and 10
Claims (3)
1. A circular cross section equal strength supporting beam receiving load evenly distributed at intervals is characterized in that: the beam bears evenly distributed loads at intervals, namely bears evenly distributed loads along the length direction of the beam per unit length at intervals of certain equal distance, the cross section of the beam is a solid circle or a hollow circle, namely a circular ring, the radius of the circle is changed along with the length direction size x of the beam, namely the radius of the cross section at the position x away from the left end of the beam is r (x); (1) when the cross section of the beam is a solid circle, the radius of the cross section of the beam in the interval section (section without external load) is
Wherein
k is the number of the beam spacer (no external load), k is odd number, a is the length of the beam spacer (no external load), L1The length of each section of the uniformly distributed load, Q is the total load (i.e. the resultant force of all distributed loads) borne by the beam, n is the number of sections of the uniformly distributed load borne by the beam, and [ sigma ]]Allowable stress for tensile and compressive of the beam material; radius of circular cross section of beam in section bearing uniformly distributed load
Wherein
k is the number of the sections of the beam bearing the evenly distributed load sections, k is an even number, a is the length of the beam spacing section (no external load section), L1The length distributed for each section of uniform load and Q is the total load borne by the beamLoad (i.e. the resultant of all distributed loads), n is the number of segments of the uniform load borne by the beam, [ sigma ]]Allowable stress for tensile and compressive of the beam material; (2) when the cross section of the beam is a circular ring, the outer radius r (x) of the circular ring meets the following structural size constraint condition:
wherein t is the width of the ring (i.e., the difference between the outer radius and the inner radius of the ring), [ sigma ]]Allowable tension and compression stress, M, for beam materialk(x) For the bending moment of the beam at the x-section, Mk(x) The following two cases are distinguished: (i) when M isk(x) When the beam spacing segment (without external load segment) bears bending moment and the segment number mark k is odd number,
wherein
a is the length of the beam spacing section (no external load section), L1The length of each section of the uniformly distributed load, Q is the total load (namely the resultant force of all the uniformly distributed loads) borne by the beam, and n is the number of sections of the uniformly distributed load borne by the beam; (ii) when M isk(x) When the beam bears the bending moment borne by the evenly distributed load sections and the section number k is an even number,
the other symbols have the same meanings as above.
2. A round cross-section constant strength support beam under evenly spaced loads according to claim 1, wherein: the beam bears loads which are evenly distributed at intervals
The symbols have the same meanings as defined in claim 1.
3. A round cross-section constant strength support beam under evenly spaced loads according to claim 1, wherein: transverse of beamWhen the cross section is a solid circular cross section, the minimum radius of the beam is
And r isminT is more than or equal to t; when the cross-section of the beam is a hollow circular cross-section (circular ring), the minimum radius of the beam is
And c isminT is more than or equal to t; wherein [ tau ] is]The shear allowable stress of the beam material is defined, and the rest symbols are as before.
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