CN111428319A - A circular cross-section equal-strength support beam subjected to uniformly spaced loads - Google Patents

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

A circular cross-section equal-strength support beam subjected to uniformly spaced loads

technical field

The invention relates to a circular cross-section equal-strength support beam subjected to evenly distributed loads at intervals, and belongs to the technical field of mechanical structure design and manufacture.

Background technique

In practice, the technical problem of girders that bear evenly spaced loads is often encountered. For example, the aluminum foil in an aluminum foil production plant needs to be supported by girders in processes such as heat treatment. Aluminum foil is produced in rolls. The aluminum foil is wound outside a hollow core, and an aluminum coil is about 300kg to 2000kg. In the heat treatment furnace, a beam passes through the core to support several aluminum foil rolls. There is a certain interval. Figure 1 is a schematic diagram of the support of aluminum foil rolls. A beam with a length of L ₂ passes through the core to support several aluminum foil rolls, which is equivalent to bearing a uniformly distributed load. The interval between each aluminum foil roll is a, which can be used in Figure 2 A mechanical model representing this type of engineering problem. The mechanical model shown in Figure 2 is suitable for any similar engineering problem of beams bearing uniformly spaced loads. In Figure 2, L ₁ is the length distributed by the uniformly distributed load, q is the concentration or linear density of the uniformly distributed load (unit is N/m or kN/m), and x is any section of the beam AB to the beam AB The distance from the left end A, V is the support reaction force.

Due to the large external load and the special mode of action of the external load, the strength and stiffness of the supporting beam have become an urgent engineering and technical problem to be solved. The traditional design mostly determines the structural size of the beam by limiting the maximum stress of the most dangerous section of the beam. The size of the entire beam is the same as the size of the most dangerous section. In this way, where the stress is small, the material does not fully play its role. It not only causes waste, but also increases the weight of the beam, which increases the cost and makes it difficult to operate and use. A more scientific and reasonable idea should be to determine a reasonable shape and size for the component at the most economical cost under the premise of ensuring safety and reliability. For the circular cross-section equal-strength support beams of the present invention that are subjected to uniformly spaced loads, if a beam with a special structural size can be invented, each cross-section of the beam is under the same strength, and the maximum stress of each cross-section is equal everywhere. , can achieve both safe and reliable, scientific and economical technical effects. According to the relevant expertise, this can be achieved, the following will make a specific analysis. In the era of increasingly developed CNC machining technology, such a structure is also easy to manufacture.

SUMMARY OF THE INVENTION

其中

k为梁间隔段(无外载荷段)的段数标号且k为奇数、a为梁间隔段(无外载荷段)的长度、L₁为每段均布载荷所分布的长度、Q为梁所承受的总载荷(即全部分布载荷的合力)、n为梁所承受的均布载荷的段数、[σ]为梁材料的拉压许用应力；梁在承受均布载荷区段的圆横截面半径

其中

k为梁承受均布载荷区段的段数标号且k为偶数、a为梁间隔段(无外载荷段)的长度、L₁为每段均布载荷所分布的长度、Q为梁所承受的总载荷(即全部分布载荷的合力)、n为梁所承受的均布载荷的段数、[σ]为梁材料的拉压许用应力；(2)当梁的横截面为圆环时，圆环外半径r(x)符合以下结构尺寸的约束条件：

其中t为圆环的宽度(即圆环外半径与内半径之差)、[σ]为梁材料的拉压许用应力、M_k(x)为梁在x截面处的弯矩，M_k(x)分以下两种情况：(i)当M_k(x)为梁间隔段(无外载荷段)所承受的弯矩且段数标号k为奇数时，

其中

a为梁间隔段(无外载荷段)的长度、L₁为每段均布载荷所分布的长度、Q为梁所承受的总载荷(即全部均布载荷的合力)、n为梁所承受的均布载荷的段数；(ii)当M_k(x)为梁在承受均布载荷区段所承受的弯矩且段数标号k为偶数时，

其余各符号意义同前。这种梁所承受的间隔均布载荷为

各符号含义同前。当梁的横截面为实心圆横截面时，梁的最小半径是

且r_min≥t；当梁的横截面为空心圆横截面(圆环)时，梁的最小半径是

且c_min≥t；其中[τ]为梁材料的剪切许用应力，其余符号同前。The technical solution adopted by the present invention to solve the technical problem is: to invent a circular cross-section equal-strength support beam subjected to spaced and evenly distributed loads, which is characterized in that: this beam is subjected to spaced and evenly distributed loads, that is, every certain equal distance Bearing a load uniformly distributed per unit length along the length of the beam, the cross section of this beam is a solid circle or a hollow circle, that is, a ring, and the radius of the circle varies with the length x of the beam, that is, at the distance x from the left end of the beam The radius of the circle at the cross section 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 (no external load section) is

in

k is the segment number of the beam interval section (no external load section) and k is an odd number, a is the length of the beam interval section (no external load section), L ₁ is the length of the uniform load distribution of each section, Q is the beam The total load (that is, the resultant force of all the distributed loads), n is the number of segments of the uniformly distributed load on the beam, [σ] is the tensile and compressive allowable stress of the beam material; the circular cross-section of the beam in the uniformly distributed load section radius

in

k is the section number of the section where the beam is subjected to uniformly distributed load and k is an even number, a is the length of the beam interval (no external load section), L1 is the length _of each section of the uniformly distributed load, and Q is the beam borne by the beam. The total load (that is, the resultant force of all distributed loads), n is the number of segments of the uniformly distributed load on the beam, and [σ] is the allowable tensile and compressive stress of the beam material; (2) When the cross-section of the beam is a ring, the circular The outer radius r(x) of the ring meets the constraints of the following structural dimensions:

where t is the width of the ring (that is, the difference between the outer radius and the inner radius of the ring), [σ] is the tensile and compressive allowable stress of the beam material, M _k (x) is the bending moment of the beam at the x section, M _k (x) is divided into the following two cases: (i) When M _k (x) is the bending moment of the beam interval section (no external load section) and the section number k is an odd number,

in

a is the length of the interval section of the beam (no external load section), L ₁ is the length of the uniformly distributed load of each section, Q is the total load borne by the beam (that is, the resultant force of all the uniformly distributed loads), and n is the beam borne by the beam (ii) When M _k (x) is the bending moment of the beam in the section under uniform load and the number of segments k is an even number,

The rest of the symbols have the same meaning as before. The spaced uniform load borne by this beam is

Each symbol has the same meaning as before. When the cross-section of the beam is a solid circular cross-section, the minimum radius of the beam is

And r _min ≥ t; when the cross section of the beam is a hollow circular cross section (ring), the minimum radius of the beam is

And c _min ≥ t; where [τ] is the allowable shear stress of the beam material, and other symbols are the same as before.

The beneficial effects of the present invention are that the strength of the beam is equal everywhere, material is saved, the structure is simplified, and the weight is reduced.

Description of drawings

Figure 1 Schematic diagram of aluminum foil roll support

Fig. 2 Mechanical model diagram of the beam subjected to uniformly spaced loads

Fig. 3 Equal-strength support beams under working conditions of the second embodiment

Fig. 4 Equal-strength support beam under working conditions of the third embodiment

In Fig. 1: 1 is a support beam, 2 is an aluminum foil roll.

In Figure 2: L ₁ is the length distributed by the uniform load, q is the concentration or linear density of the uniform load (unit is N/m or kN/m), L ₂ is the total length of the beam, x is the beam The distance from any section of AB to the left end A of beam AB, V is the reaction force of the support, Q is the total load borne by the beam (that is, the resultant force of the uniformly distributed loads on each section of the beam), and a is the interval between the aluminum foil rolls .

Detailed ways

To achieve equal strength, the characteristics and laws of external load, internal force (moment) of the beam, and the changing trends and laws of each parameter should be analyzed and studied first, and corresponding technical measures should be taken according to the research results. Since the external load of the beam involved in the present invention is special, the technical solutions and technical measures of the present invention must also be special.

As shown in Figure 2, AB represents a circular cross-section support beam subjected to evenly spaced loads. Let the full length of the beam be L ₂ , and there are n sections of uniform loads acting on it. The resultant force of the n sections of uniform loads, that is, the total load on the beam is QN, and the concentration of each section of the uniform load, that is, the linear density is q N/ m, the distribution length of each section of uniformly distributed load is L ₁ , and the interval between each section of uniformly distributed load is a; according to the relevant professional knowledge, you can obtain the inner diameter of the cross-section at the distance x from the left end of the beam (see Figure 2). The moments are as follows:

(i) Bending moment of spacer section (no external load section)

When M _k (x) is the bending moment of the beam interval section (no external load section) and the section number k is an odd number,

其余各符号意义同前。(ii)梁在承受均布载荷区段的弯矩where k is the segment number, and k is an odd number,

The rest of the symbols have the same meaning as before. (ii) The bending moment of the beam in the section under uniform load

When M _k (x) is the bending moment of the beam in the section under uniform load and the section number k is an even number,

其余各符号意义同前。where k is the segment number, and k is an even number,

The rest of the symbols have the same meaning as before.

After analysis and research, the following characteristics can be drawn:

(1) The bending moment in the beam with uniformly spaced loads has symmetry, that is, M ₁ (x)=M _2n+1 (L ₂ -x), M ₂ (x)=M _2n (L ₂ -x), M ₃ (x)=M _2n-1 (L ₂ -x), M ₄ (x)=M _2n-2 (L ₂ -x)...M _k (x)=M _2n-k+2 (L ₂ - x).

此时k为奇数，k＝n+1。(2) When the even-numbered sections are uniformly distributed (k is an odd number), the maximum bending moment is at the midpoint of the interval section in the middle of the whole beam, and is constant

At this time, k is an odd number, and k=n+1.

若

则

此时k为偶数，k＝n+1；所述

为全梁在完全均布载荷作用下的最大弯矩。(3) When subjected to evenly distributed loads in odd-numbered sections (k is an even number), the maximum bending moment is at the midpoint of the aluminum coil in the middle of the whole beam, which is

like

but

At this time k is an even number, k=n+1; the

is the maximum bending moment of the whole beam under a completely uniform load.

处，即全梁的中点，最大弯矩是(4) To sum up, when the load is uniformly distributed in n sections, the interval section has n+1 sections, a total of 2n+1 sections, and the number of bending moments is 2n+1; when n is an odd number, the maximum bending moment is the even-numbered bending moment M _k = Mn ₊₁ (that is, k=n+1), which occurs at the midpoint of the uniformly distributed load segment in the middle, that is, the midpoint of the n+1-th segment, that is,

, which is the midpoint of the full beam, the maximum bending moment is

这是众所周知的全梁受完全均布载荷(无间隔)时的最大弯矩。n为偶数时，最大弯矩是奇数号弯矩M_k＝M_n+1(即k＝n+1)，发生在第n+1个间隔段，且在整个第n+1段上为常数，也即从

到

上为常数，最大弯矩是By the way, if L ₁ =L ₂ , which means a=0, n=1, the above formula becomes

This is the known maximum bending moment for a full beam subjected to a fully uniform load (no spacing). When n is an even number, the maximum bending moment is the odd-numbered bending moment M _k =M _n+1 (ie k=n+1), which occurs at the n+1th interval and is constant throughout the n+1th segment , that is, from

arrive

is constant, the maximum bending moment is

In addition, for odd-numbered segments (interval segments) bending moments

与前述完全吻合。This is a straight line equation. Obviously, when k<n+1, _Mk increases with the increase of x; when k>n+1, _Mk decreases with the increase of x; when k=n+1, _Mk is a constant , independent of x, is

Completely consistent with the above.

For even numbers, that is, the bending moment of the uniformly distributed load segment

得This is the parabolic equation, let

have to

之内，须有to make x ₀ fall in

within, there must be

即在整个梁的中点。最大值是

若不满足k＝n+1的关系，M_k的最大值不在

内。又由于k为偶数，故n为奇数。也与前述完全吻合。n<k<n+2, since both n and k are positive integers, k=n+1, when k=n+1,

That is, at the midpoint of the entire beam. The maximum value is

If the relationship of k=n+1 is not satisfied, the maximum value of M _k is not

Inside. And since k is even, n is odd. It is also completely consistent with the above.

A similar conclusion is also obtained for the radius of the beam, that is, the beam height, namely:

For odd-numbered segments (spacer segments, i.e. no external load segments) radii

When k<n+1, _rk (x) increases with the increase of x; when k>n+1, _rk (x) decreases with the increase of x; when k=n+1, _rk (x) is a constant, independent of x, and is

For even numbers, the radius of the uniformly distributed load segment

令

得

make

have to

之内，须有n<k<n+2，由于n、k均为正整数，故有k＝n+1，当k＝n+1时，

即在整个梁的中点。若不满足k＝n+1的关系，r_k的最大值不在

内。又由于k为偶数，故n为奇数。当

时，r_k的最大值是

This is exactly the same as the stagnation point of M _k (x), so that x ₀ falls on

Inside, there must be n<k<n+2. Since both n and k are positive integers, k=n+1, when k=n+1,

That is, at the midpoint of the entire beam. If the relationship of k=n+1 is not satisfied, the maximum value of r _k is not

Inside. And since k is even, n is odd. when

When , the maximum value of r _k is

Obviously, if the size of the beam is determined by the maximum bending moment of the whole beam, the beam will be very large (see the following example), which is neither economical nor safe, because the structure is too bulky, not only a waste of manpower, material resources, financial resources, but also very difficult to operate and use. inconvenient. The technical idea of the present invention is to adapt the structural dimension of the beam to the above-mentioned special change of the internal moment, so as to achieve the technical effect of equal strength everywhere in the beam.

从而L₁＝564mm，均布载荷

根据式(1)(奇数段，即间隔段的弯矩)和式(2)(偶数段，即均布载荷段的弯矩)以及本发明的技术方案，可得各区段的弯矩M_k(x)(kNm)和梁的半径尺寸r_k(x)(m)如下：Embodiment 1, set the number of segments of the uniform load on the solid circular section beam n=5, then the maximum sequence number of odd-numbered bending moments is M _2n+1 , namely M ₁₁ , and the maximum sequence number of even-numbered bending moments is M _2n , that is, M ₁₀ ; The total load Q=73.5kN of the uniformly distributed load of the 5 sections, so the support reaction force V=Q/2=36.75kN; the beam length L ₂ =3m, the interval between the uniformly distributed loads of each section

Thus L ₁ =564mm, uniform load

According to formula (1) (odd-numbered section, namely the bending moment of the interval section) and formula (2) (even-numbered section, namely the bending moment of the uniformly distributed load section) and the technical scheme of the present invention, the bending moment M _k of each section can be obtained (x)(kNm) and the radial dimension r _k (x)(m) of the beam as follows:

The first segment 0≤x≤a, that is, 0≤x≤0.03m segment, k=1:

M ₁ (x)=Vx=Qx/2=36.75x;

即在整个第一段，每个横截面上的最大应力都恒为常数[σ]，其中W₁为第一段圆形横截面的抗弯截面系数，

can be verified,

That is, throughout the first section, the maximum stress on each cross section is constant [σ], where W ₁ is the flexural section coefficient of the circular cross section of the first section,

The second segment a≤x≤a+L ₁ , that is, 0.03m≤x≤0.594m segment, k=2:

即在整个第二段，每个横截面上的最大应力都恒为常数[σ]，其中W₂为第二段圆形横截面的抗弯截面系数，

can be verified,

That is, throughout the second section, the maximum stress on each cross section is constant [σ], where W ₂ is the flexural section coefficient of the circular cross section of the second section,

The starting point is represented by s and the end point is represented by u. It can be verified that when x=a, M _1u =M _2s .

The third segment a+L ₁ ≤x≤2a+L ₁ , that is, 0.594m≤x≤0.624m segment, k=3:

It can be verified that when x=2a+L ₁ , M _2u =M _3s .

即在整个第三段，每个横截面上的最大应力都恒为常数[σ]，其中W₃为第三段圆形横截面的抗弯截面系数，

can be verified,

That is, throughout the third section, the maximum stress on each cross section is constant [σ], where W ₃ is the bending section coefficient of the circular cross section of the third section,

即各段的每个横截面上的最大应力都恒为常数[σ]，其中

为第k段圆形横截面的抗弯截面系数。这就是等强度的优势。Similarly, it can be verified that the following paragraphs have

That is, the maximum stress at each cross section of each segment is constant [σ], where

is the flexural section coefficient of the circular cross section of the kth segment. This is the advantage of equal strength.

The fourth segment 2a+L ₁ ≤x≤2a+2L ₁ , that is, 0.624m≤x≤1.188m segment, k=4:

When x=2a+L ₁ , M _3u =M _4s ; and, M _4max =M _4u .

The fifth segment 2a+2L ₁ ≤x≤3a+2L ₁ is 1.188m≤x≤1.218m segment, k=5:

It can be verified that when x=2a+2L ₁ , M _4u =M _5s .

The sixth segment 3a+2L ₁ ≤x≤3a+3L ₁ is 1.218m≤x≤1.782m segment, k=6:

It can be verified that when x=3a+3L ₁ , M _5u =M _6s =M _5max .

得make

have to

正好是3a+2L₁与3a+3L₁中点(当n＝5时)。最大的弯矩是M₆，在

处，为

It is exactly the midpoint of 3a+2L ₁ and 3a+3L ₁ (when n=5). The maximum bending moment is M ₆ , at

at, for

M _max =M _6max =Q(0.9a+0.625L ₁ )=27.89325 kNm.

The seventh segment 3a+3L ₁ ≤x≤4a+3L ₁ is the segment 1.782m≤x≤1.812m, k=7:

The ninth segment 4a+4L ₁ ≤x≤5a+4L ₁ is the segment 2.376m≤x≤2.406m, k=9:

Section 11 5a+5L ₁ ≤x≤6a+5L ₁ ie 2.97m≤x≤3m section, k=11:

The eighth segment 4a+3L ₁ ≤x≤4a+4L ₁ is 1.812m≤x≤2.376m segment, k=8:

Section 10 5a+4L ₁ ≤x≤5a+5L ₁ is 2.406m≤x≤2.97m segment, k=10:

若取[σ]＝170000kN/m²、b＝40mm，则整个梁每一横截面的直径都必须为2r＝118.6725033mm，这正是等强度梁最高处的直径：

If the beam height is controlled by the maximum bending moment of the whole beam M _max =M _6max =27.89325kNm, the whole beam is a constant radius

If [σ]=170000kN/m ² and b=40mm, the diameter of each cross-section of the entire beam must be 2r=118.6725033mm, which is the diameter of the highest point of the beam with equal strength:

When the allowable shear stress [τ] = 90MPa, the minimum radius

The beam height 2r _k (x) of the equal-strength beam of the present invention has the above-mentioned characteristics and laws, and the beams whose shape and size conform to the above-mentioned characteristics and laws must have the same strength everywhere, that is, the maximum stress of each cross-section is equal. The same is true for the second and third embodiments, and the second and third embodiments also provide equal-strength verification.

从而L₁＝360mm，均布载荷

根据式(1)(奇数段，即间隔段的弯矩)和式(2)(偶数段，即均布载荷段的弯矩)以及本发明的技术方案，可得各区段的弯矩M_k(x)(kNm)和梁的半径r_k(x)(m)如下(2r_k(x)即为梁高)。Example 2, set the number of segments of the uniformly distributed load on the solid circular section beam n=5, then the maximum sequence number of odd-numbered bending moments is M _2n+1 , namely M ₁₁ , and the maximum sequence number of even-numbered bending moments is M _2n , that is, M ₁₀ ; The total load Q=73.5kN of the uniformly distributed load of the 5 sections, so the support reaction force V=Q/2=36.75kN; the beam length L ₂ =3m, the interval between the uniformly distributed loads of each section

Thus L ₁ =360mm, uniform load

According to formula (1) (odd-numbered section, namely the bending moment of the interval section) and formula (2) (even-numbered section, namely the bending moment of the uniformly distributed load section) and the technical scheme of the present invention, the bending moment M _k of each section can be obtained (x) (kNm) and the beam radius r _k (x) (m) are as follows (2r _k (x) is the beam height).

(The external loads of the first and second examples are exactly the same, and the expressions of the bending moment and radius are also the same, but the interval of the uniform load is different, so the length of the section suitable for the bending moment is different, which also leads to the bending moment, The size of the radius is different)

The first segment 0≤x≤a, that is, 0≤x≤0.2m segment, k=1:

M ₁ (x)=Vx=Qx/2=36.75x;

即在整个第一段，每个横截面上的最大应力都恒为常数[σ]，其中W₁为第一段圆形横截面的抗弯截面系数，

can be verified,

That is, throughout the first section, the maximum stress on each cross section is constant [σ], where W ₁ is the flexural section coefficient of the circular cross section of the first section,

The second segment a≤x≤a+L ₁ , that is, 0.2m≤x≤0.56m segment, k=2:

即在整个第二段，每个横截面上的最大应力都恒为常数[σ]，其中W₂为第二段圆横截面的抗弯截面系数，

can be verified,

That is, throughout the second segment, the maximum stress on each cross section is constant [σ], where W ₂ is the flexural section coefficient of the second circular cross section,

The starting point is represented by s and the end point is represented by u. It can be verified that when x=a, M _1u =M _2s .

The third segment a+L ₁ ≤x≤2a+L ₁ , that is, 0.56m≤x≤0.76m segment, k=3:

It can be verified that when x=2a+L ₁ , M _2u =M _3s .

即在整个第三段，每个横截面上的最大应力都恒为常数[σ]，其中W₃为第三段圆形横截面的抗弯截面系数，

can be verified,

That is, throughout the third section, the maximum stress on each cross section is constant [σ], where W ₃ is the bending section coefficient of the circular cross section of the third section,

即各段的每个横截面上的最大应力都恒为常数[σ]，其中

为第k段圆形横截面的抗弯截面系数。这就是等强度的优势。Similarly, it can be verified that the following paragraphs have

That is, the maximum stress at each cross section of each segment is constant [σ], where

is the flexural section coefficient of the circular cross section of the kth segment. This is the advantage of equal strength.

The fourth segment 2a+L ₁ ≤x≤2a+2L ₁ , that is, 0.76m≤x≤1.12m segment, k=4:

When x=2a+L ₁ , M _3u =M _4s ; and, M _4max =M _4u .

The fifth segment 2a+2L ₁ ≤x≤3a+2L ₁ is 1.12m≤x≤1.32m segment, k=5:

It can be verified that when x=2a+2L ₁ , M _4u =M _5s .

The sixth segment 3a+2L ₁ ≤x≤3a+3L ₁ is 1.32m≤x≤1.68m segment, k=6:

It can be verified that when x=3a+3L ₁ , M _5u =M _6s =M _5max .

得make

have to

正好是3a+2L₁与3a+3L₁中点(当n＝5时)。最大的弯矩是M₆，在

处，为

It is exactly the midpoint of 3a+2L ₁ and 3a+3L ₁ (when n=5). The maximum bending moment is M ₆ , at

place, for

M _max =M _6max =Q(0.9a+0.625L ₁ )=29.7675kNm

Section 7 3a+3L ₁ ≤x≤4a+3L ₁ is 1.68m≤x≤1.88m segment, k=7:

The ninth segment 4a+4L ₁ ≤x≤5a+4L ₁ is 2.24m≤x≤2.44m segment, k=9:

Section 11 5a+5L ₁ ≤x≤6a+5L ₁ ie 2.8m≤x≤3m section, k=11:

The eighth segment 4a+3L ₁ ≤x≤4a+4L ₁ is 1.88m≤x≤2.24m segment, k=8:

Section 10 5a+4L ₁ ≤x≤5a+5L ₁ is 2.44m≤x≤2.8m, k=10:

若取[σ]＝170000kN/m²，则整个梁每一横截面的直径都必须为2r＝121.2731108mm，这正是等强度梁最高处的高度：

If the beam radius is controlled by the maximum bending moment of the whole beam M _max =M _6max =29.7675kNm, the diameter of the whole beam is constant

If [σ]=170000kN/m ² , the diameter of each cross-section of the entire beam must be 2r=121.2731108mm, which is the height of the highest point of the beam with equal strength:

在梁的首末两端，即0～10mm段和2990～3000mm段取r＝14mm。When the allowable shear stress [τ] = 90MPa, the minimum radius

Take r=14mm at the beginning and end of the beam, that is, the 0-10mm section and the 2990-3000mm section.

The shape and size of the beams are determined by the above expressions and data. For the working conditions given in this embodiment, the beams with the above characteristics must be equal-strength beams. FIG. 3 is a schematic diagram of the equal-strength beam of the second embodiment.

从而L₁＝360mm，均布载荷

根据式(1)(奇数段，即间隔段的弯矩)和式(2)(偶数段，即均布载荷段的弯矩)以及本发明的技术方案与措施，可得各区段的弯矩M_k(x)(kNm)如实施例二所述。空心圆形截面(圆环)外半径r(x)符合以下结构尺寸的约束条件：The third embodiment, the parameters of each working condition are the same as the second embodiment, the hollow circular section beam is used, the number of segments with uniform load is n=5, the maximum sequence number of odd-numbered bending moments is M _2n+1 , namely M ₁₁ , and the maximum sequence number of even-numbered bending moments is M 2n+1 . is M _2n , namely M ₁₀ ; the total load Q=73.5kN of the uniformly distributed load of the 5 sections, so the support reaction force V=Q/ ₂ =36.75kN; the beam length L2=3m, the interval between the uniformly distributed loads of each section

Thus L ₁ =360mm, uniform load

According to formula (1) (odd-numbered section, namely the bending moment of the interval section) and formula (2) (even-numbered section, namely the bending moment of the uniformly distributed load section) and the technical solutions and measures of the present invention, the bending moment of each section can be obtained M _k (x) (kNm) is as described in the second embodiment. The outer radius r(x) of the hollow circular section (ring) meets the constraints of the following structural dimensions:

Where t is the peripheral width of the ring, [σ] is the allowable tensile and compressive stress of the beam material (same as in the second embodiment), and M _k (x) is the bending moment of the beam at the x section, which is exactly the same as the second embodiment. In order to demonstrate the specific implementation process, the above-mentioned constraints are further described. Transform the above formula into

D＝－0.25t⁴，Let A=t, B=-1.5t ² ,

D=-0.25t ⁴ ,

order again

Then there are:

(1) If Δ>0,

或

(2) If Δ=0 and p≥0,

or

(3) If Δ<0,

or

or

in

If the beam radius r _k (x) has the above rules and characteristics, it must be a beam of equal strength, that is, the strength of the beam is equal everywhere, that is, the maximum stress on each cross-section of the beam is the same.

且r_min≥t。以本实施例的工况条件为例，取圆环的t＝10mm，拉压许用[σ]＝170MPa，剪切许用应力[τ]＝90MPa。The reasonable and optimal beam radius can be selected according to the above rules, and the minimum radius is guaranteed.

and r _min ≥ t. Taking the working conditions of this embodiment as an example, take t=10mm of the ring, allowable tension and compression [σ]=170MPa, and allowable shear stress [τ]=90MPa.

According to the technical scheme of the present invention, the bending moments and beam heights of equal-strength beams are as follows (the bending moments and their suitable section lengths in the second and third embodiments are exactly the same, but the cross sections are different, so the beam radii of the two are the same. There are different). In order to show the dimension values, the following expressions are not given for the beam radius as in Example 2, but specific data are listed every 10 mm along the beam length, see Tables 1 to 4. In order to verify the technical effect of the present invention, several groups of data are randomly selected for inspection below.

The first segment 0≤x≤a, that is, 0≤x≤0.2m segment, k=1: M ₁ (x)=Vx=Qx/2=36.75x;

The third segment a+L ₁ ≤x≤2a+L ₁ , that is, 0.56m≤x≤0.76m segment, k=3:

The fifth segment 2a+2L ₁ ≤x≤3a+2L ₁ is 1.12m≤x≤1.32m segment, k=5:

可在梁的首末两端，即在0～30mm段和2970～3000mm段取半径r＝22mm>t，见表1和表4括号中的数据。Minimum radius

The radius r=22mm>t can be taken at the beginning and end of the beam, that is, the 0-30mm section and the 2970-3000mm section, see the data in brackets in Table 1 and Table 4.

verify:

(1) Take the data at x=100mm in the first section (see Table 1): the radius of the beam r(x=100)=33.033mm (the exact data is 33.03290622mm), the bending moment M(x)=3.675kNm, the aforementioned It has been determined that t=10mm. then

Substituting t=0.01m, r(x)=0.03303290622m, W=2.16176×10 ^-5 m ³ .

The maximum stress is

(2) Take the data at x=2000mm in Section 8 (see Table 4): the radius of the beam r(x=2000)=77.599mm (the exact data is 77.59901235mm), the bending moment M(x)=26.46kNm,

Substitute t=0.01m, r(x=2000)=0.07759901235m to obtain W=0.000155647m ³ .

The maximum stress is

The stress of other places can be calculated in the same way, and the result is that the maximum stress σ = 170MPa everywhere. This is the equal strength advantage achieved by the present invention.

Section 7 3a+3L ₁ ≤x≤4a+3L ₁ is 1.68m≤x≤1.88m segment, k=7:

The ninth segment 4a+4L ₁ ≤x≤5a+4L ₁ is 2.24m≤x≤2.44m segment, k=9:

Section 11 5a+5L ₁ ≤x≤6a+5L ₁ ie 2.8m≤x≤3m section, k=11:

The second segment a≤x≤a+L ₁ , that is, 0.2m≤x≤0.56m segment, k=2:

The fourth segment 2a+L ₁ ≤x≤2a+2L ₁ , that is, 0.76m≤x≤1.12m segment, k=4:

The sixth segment 3a+2L ₁ ≤x≤3a+3L ₁ is 1.32m≤x≤1.68m segment, k=6:

The eighth segment 4a+3L ₁ ≤x≤4a+4L ₁ is 1.88m≤x≤2.24m segment, k=8:

Section 10 5a+4L ₁ ≤x≤5a+5L ₁ is 2.44m≤x≤2.8m, k=10:

With the data in Tables 1 to 4, the specific size and shape of the equal-strength beams can be determined. The equal-strength beam obtained from the working conditions of Example 3 and the technical solution of the present invention is shown in FIG. 4 . Compared with Example 2, the beam height of Example 3 is greater than that of Example 2, but since the beam of Example 3 is hollow, its cross-sectional area is smaller, which can be verified by using the beam height data in Tables 1 to 4. at this point.

Table 1 Beam radius and bending moment values for the 1st, 3rd and 5th sections

Table 2 Beam radius and bending moment values of the 7th, 9th and 11th sections

Table 3 Beam radius and bending moment values for the second, fourth and sixth sections

Table 4 Beam radius and bending moment values in 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 material_k(x) For the bending moment of the beam at the x-section, M_k(x) The following two cases are distinguished: (i) when M is_k(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 is_k(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 is_minT 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 is_minT 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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