CN112287441B - Method for calculating influence line of uniform-span uniform-section continuous beam - Google Patents

Method for calculating influence line of uniform-span uniform-section continuous beam Download PDF

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CN112287441B
CN112287441B CN202011189123.6A CN202011189123A CN112287441B CN 112287441 B CN112287441 B CN 112287441B CN 202011189123 A CN202011189123 A CN 202011189123A CN 112287441 B CN112287441 B CN 112287441B
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周毅
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University of Science and Technology Beijing USTB
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Abstract

The invention provides a method for calculating influence lines of an equal-span uniform-section continuous beam, and belongs to the technical field of structural analysis. The method comprises the steps of firstly determining a solution of the multi-span continuous beam when the leftmost beam end is subjected to the action of concentrated bending moment, then determining the solution of the multi-span continuous beam when the rightmost beam end is subjected to the action of concentrated bending moment according to a mirror image relationship, then determining the beam end rotating rigidity of the multi-span continuous beam, determining a node bending moment influence line of the multi-span continuous beam according to the mutual theorem of work equivalence, and finally determining an analytical formula of other influence lines. According to the method, aiming at the continuous beam with equal span and arbitrary span number and uniform section, an influence line analytical formula of internal force, displacement and support reaction of the structure and a general calculation formula of the structural response of the continuous beam are deduced by introducing the beam end rotating rigidity of the multi-span beam. The formula belongs to an accurate solution, is convenient to calculate, has strong universality and is convenient for parameter analysis; the limit value of the structural response when the span number of the continuous beam tends to infinity can be given, and the mechanical characteristics of the continuous beam are revealed in a panoramic way.

Description

Method for calculating influence line of uniform-span uniform-section continuous beam
Technical Field
The invention relates to the technical field of structural analysis, in particular to a method for calculating influence lines of an equal-span equal-section continuous beam.
Background
The continuous beam is widely applied to projects such as buildings, bridges, pipelines and the like. Although the continuous beam has various forms, the most classical structural form is the constant-span constant-section continuous beam, and the solution method of the constant-span constant-section continuous beam is widely researched. In addition to the finite element analysis method, the theoretical solution method includes a classical force method (based on a three-bending moment equation), a displacement method, a moment distribution method, a cubic spline beam function method, a direct writing method, a finite difference method, an equal straight beam displacement method, a piecewise independent integral method, a distributed transfer function method and the like. The methods have the characteristics and pertinence, but in terms of the calculation of the equal-span constant-section continuous beam, the steps are more, and the application is inconvenient. More importantly, the existing method is only suitable for the continuous beam with the specified span, and the change rule of the deformation and the internal force of the continuous beam along with the span is not easy to obtain. A unified calculation formula which is simple in form and clear in parameter relation and can freely span continuous beams is not seen so far.
The influence line is independent of the specific load and is an inherent property of the continuous beam. The method has the advantages that a calculation formula of influence lines of the equal-span and equal-section continuous beam is established, so that the stress characteristics of the continuous beam can be disclosed, and a theoretical basis is provided for design and monitoring of the continuous beam.
Disclosure of Invention
The invention aims to provide a method for calculating influence lines of an equal-span equal-section continuous beam.
The method comprises the following steps:
(1) Determining the solution of the multi-span continuous beam when the leftmost beam end is subjected to the action of concentrated bending moment:
for n-span equal-section continuous beam, the bending moment M is applied to the leftmost end C Angle of rotation z acting at node j j,0 Comprises the following steps:
Figure BDA0002752277880000021
wherein: n is the total span of the continuous beam; j is a node number, and j =0,1, \8230, n is sequentially arranged from left to right; z is a radical of j,0 Is the angle of rotation of the node j, clockwise rotation being positive, 0 in the subscript representing the bending moment M C Acting on node 0; m is a group of C The beam end concentrates bending moment, so that the tension of the lower edge of the main beam is positive; i all right angle 0 Is the bending linear stiffness per span of the continuous beam, defined as i 0 ＝EI/l 0 (ii) a Wherein E is the elastic modulus of the main beam material, I is the bending moment of inertia of the main beam section, and l 0 Is the span of each span;
Figure BDA0002752277880000022
wherein constant is
Figure BDA0002752277880000023
Figure BDA0002752277880000024
Figure BDA0002752277880000025
And
Figure BDA0002752277880000026
the superscripts n-j and n of (a) represent the indices;
bending moment M C Moment M acting on the leftmost end of the continuous beam at node j j,0 Comprises the following steps:
Figure BDA0002752277880000027
wherein: m j,0 Is the moment at node j so that the lower edge of the main beam is pulled positively, and 0 in the subscript represents moment M C Acts on node 0;
Figure BDA0002752277880000028
bending moment M C Acting on the leftmost end of the continuous beam at x = (k + xi) l 0 Vertical displacement at position D k+ξ,0 Comprises the following steps:
Figure BDA0002752277880000029
wherein: d k+ξ,0 Is positioned in the k +1 midspan xi l 0 The vertical displacement at the position takes the vertical downward displacement as positive, wherein k =0,1, \8230, n-1,0 ≦ ξ ≦ 1; in the subscripts, 0 represents a bending moment M C Acting on node 0; g 1 ＝ξ(1-ξ) 2 ，g 2 ＝ξ 2 (1-ξ)；
Figure BDA00027522778800000210
Figure BDA00027522778800000211
And
Figure BDA00027522778800000212
the superscripts n-k and n-k-1 of (A) represent indices;
(2) Determining the solution of the multi-span continuous beam when the rightmost beam end is subjected to the action of concentrated bending moment according to the mirror image relationship:
for the n-span equal-section continuous beam, the rightmost end of the beam is subjected to bending moment M C Angle of rotation z acting at node j j,n Bending moment M j,n And when x = (k + ξ) l 0 Vertical displacement at position D k+ξ,n Respectively as follows:
Figure BDA00027522778800000213
Figure BDA00027522778800000214
Figure BDA0002752277880000031
wherein: in the subscripts, n represents a bending moment M C Acting on a node n;
Figure BDA0002752277880000032
Figure BDA0002752277880000033
Figure BDA0002752277880000034
and
Figure BDA0002752277880000035
superscripts j, k and k +1 of (a) represent indices;
(3) Determining the beam end rotational stiffness of the multi-span continuous beam:
for the n-span equal-section continuous beam, the bending moment required to be applied when the leftmost end generates a unit corner is defined as the beam end rotating rigidity K n From step (1) K can be obtained n The calculation formula of (2) is as follows:
Figure BDA0002752277880000036
wherein: k n Is the rotational stiffness of the beam end, so that the tension of the lower edge of the main beam is positive,
Figure BDA0002752277880000037
according to the mirror image relationship, the bending moment required to be applied when the unit corner occurs at the rightmost end of the n-span continuous beam is-K n ；
(4) Determining a node bending moment influence line of the multi-span continuous beam according to the mutual equality theorem of work:
for the n-span equal-span uniform-section continuous beam, the bending moment influence line of the n-span equal-section continuous beam at the joint j is marked as M j (x) J =0,1, \8230, n, then for x = (k + xi) l 0 There are the following calculation formulas:
Figure BDA0002752277880000038
(5) Determining analytical formulas of other influence lines according to the node bending moment influence lines in the step (4):
n span equal section continuous beam is y = (r + lambda) l 0 The influence of the bending moment at the location is marked M r+λ (x) Where r =0,1, \8230, n-1,0 ≦ λ ≦ 1, and the argument x = (k + ξ) l 0 ，k＝0,1,…,n-1，0≤ξ≤1；M r+λ (x) The expression of (a) is:
Figure BDA0002752277880000039
wherein:
Figure BDA0002752277880000041
Figure BDA0002752277880000042
Figure BDA0002752277880000043
Figure BDA0002752277880000044
and
Figure BDA0002752277880000045
the superscript n-r,n-r-1, r and r +1 represent indices;
n-span equal-section continuous beam is y = (r + lambda) l 0 The shear force influence line at the location is denoted Q r+λ (x) If the shear force is set so that the beam segment rotates clockwise as positive, then:
Figure BDA0002752277880000046
wherein:
Figure BDA0002752277880000047
shear force influence line Q r+λ (x) At x = (r + λ) l 0 The presence of a mutation;
the reaction influence line of the n-span uniform-section continuous beam support p is denoted as R p (x) Where p =0,1, \8230;, n, with the counter force being positive upwards, then:
when p = 0:
Figure BDA0002752277880000048
wherein,
Figure BDA0002752277880000049
Figure BDA00027522778800000410
and
Figure BDA00027522778800000411
the superscripts n-1 and 1 of (a) represent the index;
when p =1,2, \8230;, n-1:
Figure BDA0002752277880000051
wherein:
Figure BDA0002752277880000052
Figure BDA0002752277880000053
Figure BDA0002752277880000054
Figure BDA0002752277880000055
Figure BDA0002752277880000056
Figure BDA0002752277880000057
and
Figure BDA0002752277880000058
the superscripts p-1, p, n-p-1, n-p and p +1 denote indices;
when p = n:
Figure BDA0002752277880000059
n span equal section continuous beam is y = (r + lambda) l 0 The influence line of the displacement at the position is denoted as D r+λ (x) The expression is as follows:
Figure BDA00027522778800000510
wherein f is 1 ＝λ(1-λ)(1+λ)，f 2 λ (1- λ) (2- λ), and:
Figure BDA0002752277880000061
n-span equal-section continuous beam at y =(r+λ)l 0 The influence line of the rotation angle at the position is represented as theta r+λ (x) The expression is as follows:
Figure BDA0002752277880000062
wherein, f' 1 ＝1-3λ 2 ，f′ 2 ＝3λ 2 -6 λ +2, and:
Figure BDA0002752277880000063
the continuous beam is a straight continuous beam with equal cross sections and the span and the material are the same, and the shear deformation is not considered in the analysis.
Deducing a corner analytical formula of each node when the n-span equal-section continuous beam is subjected to load action on the u-th span according to the beam end rotational stiffness in the step (3):
Figure BDA0002752277880000064
wherein: n is a positive integer, u =1,2, \ 8230;, n; j is a node number, and j =0,1, \ 8230;, n; z is a radical of formula j Is the corner of node j, and clockwise rotation is positive;
Figure BDA0002752277880000065
and
Figure BDA0002752277880000066
respectively, the u-th span main beam has fixed end bending moment under the action of load, and the tension of the lower edge of the section is taken as positive;
Figure BDA0002752277880000067
Figure BDA0002752277880000068
Figure BDA0002752277880000069
and
Figure BDA00027522778800000610
the superscripts n-u +1, n-u, u-1 and u denote indices;
according to the equation of angular displacement in structural mechanics, from z j And obtaining the bending moment of each node of the main beam, and further calculating the internal force and displacement of any section position.
When the n-span equal-section continuous beam is under the load action on the u-span, and the fixed end bending moment generated by the load meets the requirement
Figure BDA0002752277880000071
When the condition is satisfied, the rotation angle z of each node is j Bending moment M j The analytical formulas are respectively as follows:
Figure BDA0002752277880000072
Figure BDA0002752277880000073
wherein,
Figure BDA0002752277880000074
when the load of each span of the n-span equal-section continuous beam is the same and the fixed end bending moment generated by the load form meets the requirement
Figure BDA0002752277880000075
In the condition (2), all node rotation angles z 'after each cross-loading effect are considered' j And bending moment M' j The analytical formulas of (a) are respectively as follows:
Figure BDA0002752277880000076
Figure BDA0002752277880000077
the technical scheme of the invention has the following beneficial effects:
in the scheme, the given analytic formula belongs to an accurate solution, the calculation is convenient, the complex structural mechanics analysis is avoided, the finite element numerical simulation is also avoided, the method is very suitable for field calculation and test result check, and the method has obvious practical application value. More importantly, the analytic formula provided by the invention is suitable for the uniform-span uniform-section continuous beam with any span number, has strong universality and is convenient for parameter analysis; the method can provide the limit value of structural response when the span of the continuous beam tends to infinity, reveals the mechanical characteristics of the continuous beam in a panoramic way, and has definite theoretical significance. In addition, the result obtained by the invention can provide a basis for the design and monitoring of the continuous beam.
Drawings
FIG. 1 is an analysis model of the effect of concentrated bending moment on the leftmost end of a continuous beam in an embodiment of the invention;
FIG. 2 is an analysis model for solving the bending moment influence line of the continuous beam joint in the embodiment of the invention;
FIG. 3 is an analysis model for solving any section bending moment influence line of the continuous beam in the embodiment of the invention, wherein (a) the r +1 st span main beam is subjected to unit load and beam end bending moment, (b) the r +1 st span main beam is only subjected to unit load, and (c) the r +1 st span main beam is only subjected to beam end bending moment;
FIG. 4 is an analysis model of the continuous beam when the u-th span is under load in the embodiment of the invention.
Detailed Description
In order to make the technical problems, technical solutions and advantages of the present invention more apparent, the following detailed description is given with reference to the accompanying drawings and specific embodiments.
The invention provides a method for calculating influence lines of an equal-span constant-section continuous beam.
The method comprises the following steps:
(1) Determining the solution of the multi-span continuous beam when the leftmost beam end is subjected to the action of concentrated bending moment:
for n-span equal-section continuous beam, the bending moment M is applied to the leftmost end C Angle of rotation z acting at node j j,0 Comprises the following steps:
Figure BDA0002752277880000081
wherein: n is the total span of the continuous beam; j is a node number, and j =0,1, \ 8230;, n; z is a radical of formula j,0 Is the angle of rotation of the node j, clockwise rotation being positive, 0 in the subscript representing the bending moment M C Acting on node 0; m C The beam end concentrates bending moment, so that the tension of the lower edge of the main beam is positive; i.e. i 0 Is the bending line stiffness per span of the continuous beam, defined as i 0 ＝EI/l 0 (ii) a Wherein E is the elastic modulus of the main beam material, I is the bending moment of inertia of the main beam section, and l 0 Is the span of each span;
Figure BDA0002752277880000082
wherein constant is
Figure BDA0002752277880000083
Figure BDA0002752277880000084
Figure BDA0002752277880000085
And
Figure BDA0002752277880000086
the superscripts n-j and n of (a) represent the indices;
bending moment M C Bending moment M acting at joint j at the leftmost end of continuous beam j,0 Comprises the following steps:
Figure BDA0002752277880000087
wherein: m is a group of j,0 Is the moment at node j so that the lower edge of the main beam is pulled positively, and 0 in the subscript represents moment M C Acting on node 0;
Figure BDA0002752277880000091
bending moment M C Acting on the leftmost end of the continuous beam at x = (k + xi) l 0 Vertical displacement at position D k+ξ,0 Comprises the following steps:
Figure BDA0002752277880000092
wherein: d k+ξ,0 Is located in the k +1 th span xi l 0 The vertical displacement at the position takes the vertical downward displacement as positive, wherein k =0,1, \8230, n-1,0 is less than or equal to ξ is less than or equal to 1; in the subscripts, 0 represents a bending moment M C Acts on node 0; g 1 ＝ξ(1-ξ) 2 ，g 2 ＝ξ 2 (1-ξ)；
Figure BDA0002752277880000093
Figure BDA0002752277880000094
And
Figure BDA0002752277880000095
the superscripts n-k and n-k-1 of (A) represent indices;
(2) According to the mirror image relationship, determining the solution of the multi-span continuous beam when the rightmost beam end is subjected to the action of concentrated bending moment:
for the n-span equal-section continuous beam, the rightmost end of the beam is subjected to bending moment M C Angle of rotation z acting at node j j,n Bending moment M j,n And when x = (k + ξ) l 0 Vertical displacement at position D k+ξ,n Respectively as follows:
Figure BDA0002752277880000096
Figure BDA0002752277880000097
Figure BDA0002752277880000098
wherein: in the subscripts, n represents a bending moment M C Acting on the node n;
Figure BDA0002752277880000099
Figure BDA00027522778800000910
Figure BDA00027522778800000911
and
Figure BDA00027522778800000912
the superscripts j, k and k +1 of (a) represent the exponents;
(3) Determining the beam end rotational stiffness of the multi-span continuous beam:
for an n-span equal-span uniform-section continuous beam, the bending moment required to be applied when the leftmost end generates a unit corner is defined as the beam end rotating rigidity K n From step (1) K can be obtained n The calculation formula of (c) is:
Figure BDA00027522778800000913
wherein: k is n Is the rotational stiffness of the beam end, so that the tension of the lower edge of the main beam is positive,
Figure BDA00027522778800000914
according to the mirror image relationship, the bending moment required to be applied when the unit corner occurs at the rightmost end of the n-span continuous beam is-K n ；
(4) Determining a node bending moment influence line of the multi-span continuous beam according to the mutual theorem of work:
for the n-span equal-span uniform-section continuous beam, the bending moment influence line of the n-span equal-section continuous beam at the joint j is marked as M j (x) J =0,1, \8230, n, then for x = (k + xi) l 0 There is a calculation formula:
Figure BDA0002752277880000101
(5) Determining analytical formulas of other influence lines according to the node bending moment influence lines in the step (4):
n-span equal-section continuous beam is y = (r + lambda) l 0 The influence of the bending moment at the location is marked M r+λ (x) Where r =0,1, \8230, n-1,0 ≦ λ ≦ 1, and the argument x = (k + ξ) l 0 ，k＝0,1,…,n-1，0≤ξ≤1；M r+λ (x) The expression of (a) is:
Figure BDA0002752277880000102
wherein:
Figure BDA0002752277880000103
Figure BDA0002752277880000104
Figure BDA0002752277880000105
Figure BDA0002752277880000106
and
Figure BDA0002752277880000107
the superscripts n-r, n-r-1, r and r +1 of (A) represent indices;
n span equal section continuous beam is y = (r + lambda) l 0 The shear force influence line at the location is denoted Q r+λ (x) Setting the shear force to make the beam segment rotate clockwise as positive, then:
Figure BDA0002752277880000111
wherein:
Figure BDA0002752277880000112
shear force influence line Q r+λ (x) At x = (r + λ) l 0 The presence of a mutation;
the reaction force influence line of the n-span equal-span uniform-section continuous beam support p is denoted as R p (x) Where p =0,1, \8230;, n, with the counter force being positive upwards, then:
when p = 0:
Figure BDA0002752277880000113
wherein,
Figure BDA0002752277880000114
Figure BDA0002752277880000115
and
Figure BDA0002752277880000116
the superscripts n-1 and 1 of (a) represent the index;
when p =1,2, \8230;, n-1:
Figure BDA0002752277880000117
wherein:
Figure BDA0002752277880000118
Figure BDA0002752277880000121
Figure BDA0002752277880000122
Figure BDA0002752277880000123
Figure BDA0002752277880000124
Figure BDA0002752277880000125
and
Figure BDA0002752277880000126
the superscripts p-1, p, n-p-1, n-p and p +1 denote indices;
when p = n:
Figure BDA0002752277880000127
n span equal section continuous beam is y = (r + lambda) l 0 The influence line of the displacement at the position is denoted as D r+λ (x) The expression is as follows:
Figure BDA0002752277880000128
wherein, f 1 ＝λ(1-λ)(1+λ)，f 2 λ (1- λ) (2- λ), and:
Figure BDA0002752277880000129
n span equal section continuous beam is y = (r + lambda) l 0 The influence line of the rotation angle at the position is represented as theta r+λ (x) The expression is as follows:
Figure BDA0002752277880000131
wherein, f' 1 ＝1-3λ 2 ，f′ 2 ＝3λ 2 -6 λ +2, and:
Figure BDA0002752277880000132
the continuous beam is a straight continuous beam with equal cross sections and the span and the material are the same, and the shear deformation is not considered in the analysis.
Deducing a corner analytical formula of each node when the n-span equal-span uniform-section continuous beam is under the load action of the u-span according to the beam end rotational stiffness in the step (3):
Figure BDA0002752277880000133
wherein: n is a positive integer, u =1,2, \ 8230;, n; j is a node number, and j =0,1, \ 8230;, n; z is a radical of j Is the corner of node j, and clockwise rotation is positive;
Figure BDA0002752277880000134
and
Figure BDA0002752277880000135
respectively setting the fixed end bending moment of the u-th span main beam under the action of load, and taking the tension of the lower edge of the section as positive;
Figure BDA0002752277880000136
Figure BDA0002752277880000137
Figure BDA0002752277880000138
and
Figure BDA0002752277880000139
the superscripts n-u +1, n-u, u-1 and u denote indices;
according to the equation of angular displacement in structural mechanics, from z j And obtaining the bending moment of each node of the main beam, and further calculating the internal force and displacement of any section position.
When the n-span equal-section continuous beam is in the second placeu span is under the load effect, and the fixed end bending moment generated by the load meets the requirement
Figure BDA00027522778800001310
When the condition is satisfied, the rotation angle z of each node is j Bending moment M j The analytical formulas of (a) are respectively as follows:
Figure BDA0002752277880000141
Figure BDA0002752277880000142
wherein,
Figure BDA0002752277880000143
when the loads of all the spans of the n-span equal-span uniform-section continuous beam are the same and the fixed end bending moment generated by the load form meets the requirement
Figure BDA0002752277880000144
When the condition (2) is satisfied, all the node rotation angles z 'after the respective cross-loading effects are considered' j And bending moment M' j The analytical formulas are respectively as follows:
Figure BDA0002752277880000145
Figure BDA0002752277880000146
the following description is given with reference to specific embodiments.
The derivation of the solution of the multi-span continuous beam when the leftmost beam end is subjected to the action of the concentrated bending moment in the step (1) is determined as follows:
for the n-span equal-section continuous beam in the attached figure 1, the left end of the beam is subjected to bending moment M C Under the action condition, the bending moment is set to take the tension of the lower edge of the main beam as positive, and shear is not consideredAnd (4) shaping. The rigidity of each section of main beam in bending resistance line is i 0 ＝EI/l 0 Wherein E is the elastic modulus of the material, I is the bending moment of inertia of the cross section, l 0 Is the length of each span. According to a displacement method, setting the node rotation angle of the main beam at each support as an unknown quantity, and recording as z j Wherein j =0,1, \8230, n corresponds to n +1 supports from left to right respectively, and the node rotation angle is positive by clockwise rotation.
And listing the bending moment balance equation at each node. At z 0 Treating:
4i 0 ·z 0 +2i 0 ·z 1 ＝M C (1)
at z j (j =1,2, \8230;, n-1):
2i 0 ·z j-1 +8i 0 ·z j +2i 0 ·z j+1 ＝0 (2)
at z n Treating:
2i 0 ·z n-1 +4i 0 ·z n ＝0 (3)
from formula (3):
Figure BDA0002752277880000151
let a 1 =1/2, and z is n ＝-a 1 ·z n-1 By substitution of formula (2) to give z j And z j-1 Relationship between (j = n-1, n-2, \8230; 1):
z j ＝-a n+1-j ·z j-1 (5)
wherein the array a j (j =2,3, \8230;, n) satisfies:
Figure BDA0002752277880000152
to find a j The general term of (1) is obtained by subtracting constants from both sides of the equation (6) at the same time
Figure BDA0002752277880000153
Namely, it is
Figure BDA0002752277880000154
And (3) performing equal deformation on the right side of the formula (7):
Figure BDA0002752277880000155
in the molecule
Figure BDA0002752277880000156
Can obtain the product
Figure BDA0002752277880000157
The following two formulae are obtained from formula (7):
Figure BDA0002752277880000158
Figure BDA0002752277880000159
the above two formulas are divided to obtain:
Figure BDA0002752277880000161
thus array of numbers
Figure BDA0002752277880000162
Is to use
Figure BDA0002752277880000163
Is an equal ratio series of a common ratio, the first term is
Figure BDA0002752277880000164
Thus, the device
Figure BDA0002752277880000165
Known as a 1 (ii) =1/2, and a can be determined from equation (12) j (ii) expression (j =1,2, \8230;, n):
Figure BDA0002752277880000166
will z 1 ＝-a n ·z 0 (formula (5)) into formula (1) can solve z 0 ：
Figure BDA0002752277880000167
According to the recursion relational expression (5), the rotation angle z of any node can be obtained j The calculation formula of (2). To distinguish concentrated bending moment M C Position of action, introducing the symbol z j,0 In the subscript, 0 represents a bending moment M C Acting on node 0, i.e.
Figure BDA0002752277880000168
Formula (15) applies to j =0,1, \8230, n, n is a positive integer. For simplicity of presentation, the sequence of numbers is introduced
Figure BDA0002752277880000169
And
Figure BDA00027522778800001610
they have the general term
Figure BDA00027522778800001611
Wherein m represents
Figure BDA00027522778800001612
And
Figure BDA00027522778800001613
is used as an index of (1). Accordingly, equation (15) can be expressed as:
Figure BDA00027522778800001614
according to M j,0 ＝4i 0 ·z j,0 +2i 0 ·z j+1,0 Calculating the bending moment of the joint j (taking the tension of the lower edge of the section as positive):
Figure BDA00027522778800001615
equation (17) applies to j =0,1, \8230, n.
In x = (k + xi) l 0 Vertical displacement at the location may be according to M k,0 And M k+1,0 Obtaining the parameters k =0,1, \8230, n-1,0 ≦ ξ ≦ 1. If the vertical downward displacement is taken as positive and the shear deformation is ignored, the graph multiplication can obtain:
Figure BDA0002752277880000171
wherein: d k+ξ,0 Is located in xi l in the kth span 0 Vertical displacement at position, in subscripts, 0 denotes bending moment M C Acting on node 0. Function g 1 ＝ξ(1-ξ) 2 ，g 2 ＝ξ 2 (1-ξ)。
Determining the solution of the multi-span continuous beam when the rightmost beam end of the multi-span continuous beam is subjected to the action of concentrated bending moment in the step (2), and obtaining the solution according to a mirror image relationship:
concentrated bending moment M acting on node n C Under the action of (2), the corner z of the node j j,n Bending moment M j,n And in x = (k + ξ) l 0 Vertical displacement at position D k+ξ,n Comprises the following steps:
Figure BDA0002752277880000172
Figure BDA0002752277880000173
Figure BDA0002752277880000174
wherein: in the subscripts, n represents a bending moment M C Acting on node n. Other definitions of the positive and negative values of the turning angle and the bending moment and the parameters are the same as in the step (1).
And (3) determining the specific derivation of the beam end rotational stiffness of the multi-span continuous beam as follows:
let j =0,z j,0 The formula (16) is substituted by =1, and the bending moment M required to be applied when the unit corner is generated at the leftmost end of the n-span equal-span continuous beam can be obtained C I.e. the rotational stiffness K n ：
Figure BDA0002752277880000175
According to the mirror image relationship, the rotating rigidity corresponding to the rightmost end of the n-span continuous beam is-K n 。
In the step (4), according to the mutual equivalence theorem of work, the concrete derivation of the node bending moment influence line of the multi-span continuous beam is determined as follows:
considering the bending moment influence line of the n-span equal-section continuous beam at the joint j, namely solving the external load F x =1 moment of bending M at node j under action of 1 j . According to the theorem of mutual equivalence of work, the bending moment influence line M of the joint j j (x) This is equivalent to determining the deflection curve δ of the structure of fig. 2 when a unit relative rotation (relative rotation angle θ = 1) occurs at the node j xj (x) And is provided with
M j (x)＝-δ xj (x) (23)
Suppose that the bending moment at node j in FIG. 2 is M C And the rightmost end of the j-span continuous beam on the left side of the node j has the rotational rigidity of-K j The rotational stiffness of the leftmost end of the n-j span continuous beam on the right side of the node j is K n-j Then there is
Figure BDA0002752277880000181
According to the calculation formula of the rotational stiffness in the step (3), the calculation formula can be obtained by the formula (24)
Figure BDA0002752277880000182
The formula (25) is substituted for the formulas (18) and (21), and the deflection curve delta of the n-span continuous beam can be obtained xj (x) Further obtaining the bending moment influence line M of the node j j (x) In that respect Let the argument x = (k + xi) l 0 Wherein, the parameter k =0,1, \8230, n-1,0 ≤ ξ ≤ 1, then:
Figure BDA0002752277880000183
the above formula applies to node j =0,1, \8230;, n.
And (5) determining analytical formulas of other influence lines according to the node bending moment influence lines in the step (4). The method comprises the following specific steps:
considering that the n-span equal-section continuous beam is y = (r + lambda) l 0 Influence line M of bending moment at position r+λ (x) Wherein r =0,1, \8230, n-1,0 ≤ λ ≤ 1. It can be obtained by the superposition of bending moment influence lines of the nodes r and r + 1. When unit load acts on the (r + 1) th span, the stress condition of the span is equivalent to the simply supported beam, M, shown in the attached figure 3 (a) r+λ (x) Can be represented by M according to the equilibrium relationship of FIGS. 3 (b) and (c) r+1 (x) And M r (x) Thus obtaining the product. When unit load acts beyond the r +1 th span, M r+λ (x) This can be obtained from the equilibrium relationship of FIG. 3 (c). Thus:
Figure BDA0002752277880000184
wherein,
Figure BDA0002752277880000185
the expression of (a) is:
Figure BDA0002752277880000191
m can be obtained by substituting formula (26) for formula (27) r+λ (x) In the expanded form:
Figure BDA0002752277880000192
wherein:
Figure BDA0002752277880000193
in the same way, the continuous beam with n spans, equal spans and equal sections at y = (r + lambda) l can be obtained 0 Shear force influence line Q at position r+λ (x) .1. The If the shear force is positive to rotate the beam segment clockwise, then:
Figure BDA0002752277880000194
wherein,
Figure BDA0002752277880000195
the expression of (a) is:
Figure BDA0002752277880000196
note shear force influence line Q r+λ (x) At x = (r + λ) l 0 There is a mutation. Q in formula (31) can be obtained from formula (26) r+λ (x) In the expanded form:
Figure BDA0002752277880000201
wherein:
Figure BDA0002752277880000202
the reaction force influence line of the n-span equal-span uniform-section continuous beam support p is denoted as R p (x) Wherein p =0,1, \8230;, n. R is p (x) Can be obtained by the shear balance condition between the support p and the adjacent beam section. If the reaction force is positive in the upward direction and represents the expression for the bending moment of the node in equation (26), then:
when p = 0:
Figure BDA0002752277880000203
when p =1,2, \8230;, n-1:
Figure BDA0002752277880000204
wherein:
Figure BDA0002752277880000205
Figure BDA0002752277880000211
when p = n:
Figure BDA0002752277880000212
n span equal section continuous beam is y = (r + lambda) l 0 The influence line of the displacement at the position is denoted as D r+λ (x) It can be obtained by multiplication of graphs from the bending moment influence lines of the nodes r and r + 1. When a unit load acts on the (r + 1) th span, x = (r + λ) l 0 The displacement at position can be decomposed into a superposition of the two cases of fig. 3 (b) and (c); when the unit load acts outside the r +1 th span, D r+λ (x) Can be calculated according to fig. 3 (c). Thus:
Figure BDA0002752277880000213
wherein,
Figure BDA0002752277880000214
the expression of (c) is:
Figure BDA0002752277880000215
if introducing a function f 1 ＝λ(1-λ)(1+λ)，f 2 If the expression for the bending moment of the node in equation (26) is substituted for equation (40) in terms of λ (1- λ) (2- λ), then:
Figure BDA0002752277880000216
Figure BDA0002752277880000221
wherein:
Figure BDA0002752277880000222
in formula (40), for λ l 0 Derivation is carried out, and the continuous beam with n spans, equal spans and equal sections can be obtained in the condition that y = (r + lambda) l 0 Angle of rotation influence line theta at position r+λ (x)：
Figure BDA0002752277880000223
Wherein, the function f' 1 ＝1-3λ 2 ，f′ 2 ＝3λ 2 -6 λ +2, and:
Figure BDA0002752277880000224
according to the beam end rotating rigidity in the step (3), the n-span equal-cross section in the attached figure 4 can be deducedAnd when the u-th span of the continuous beam is under the load action, the corner analytical formula of each node is obtained. According to the displacement method, the rotation angle of the nodes u-1 and u can be set as an unknown quantity x 1 And x 2 And a node bending moment balance equation is listed:
Figure BDA0002752277880000225
in the formula
Figure BDA0002752277880000226
And
Figure BDA0002752277880000227
respectively setting the fixed end bending moment of the u-th span main beam under the action of load, and taking the tension of the lower edge of the section as positive; for common loading forms, the fixed end bending moment can be obtained by looking up a table. K is u-1 And K n-u And represents the rotational stiffness of the leftmost end of the u-1 span and the n-u span respectively (formula (22)). The formulation (46) yields a simple form of a system of linear equations in two dimensions:
Figure BDA0002752277880000231
thus solving the following:
Figure BDA0002752277880000232
the rotation angles of all nodes can be obtained from the proportional relationship between the rotation angles (as shown in equations (15) and (19)):
Figure BDA0002752277880000233
according to a corner displacement equation in structural mechanics, bending moment at each node of the main beam and internal force and displacement at any cross section position can be further obtained by the formula (49).
When the fixed end bending moment is satisfied
Figure BDA0002752277880000234
Under the conditions of (1), the formula (49) can be simplified to
Figure BDA0002752277880000235
Further obtaining the bending moment M of each node j The analytical formula of (1) is as follows:
Figure BDA0002752277880000236
when the loads of all the spans of the n-span equal-span uniform-section continuous beam are the same and the fixed end bending moment generated by the load form meets the requirement
Figure BDA0002752277880000241
And when the condition is met, the node corner and the bending moment can be obtained through the superposition principle. Let the corner and the bending moment of the node j of the n-span continuous beam caused by the u-th span load be respectively recorded as z ju And M ju Consider all node turns and bending moments (denoted as z ') after each cross-load effect' j And M' j ) Is composed of
Figure BDA0002752277880000242
Figure BDA0002752277880000243
Example 1
Assuming a certain 3-span continuous beam, the span per span is l 0 And the bending rigidity EI of the main beam along the length direction is constant. The bending moment influence line equation of the middle cross-middle section is determined.
From the above conditions, span number n =3, and the intermediate span middle cross section is located at y = (1 + 1/2) l 0 Here, therefore r =1, λ =1/2. Separately determining constants
Figure BDA0002752277880000244
Figure BDA0002752277880000245
From these results, the bending moment influence line M of the mid-span cross-section can be obtained from the equation (29) 1+1/2 (x) Comprises the following steps:
Figure BDA0002752277880000246
example 2
And (3) setting the span number of the continuous beam bridge as n, and solving a node corner and a bending moment when the u-th span respectively acts on the following three loads: uniformly distributing a load q (the q is positive downwards); (2) a mid-span concentration force F (F is positive going down); (3) The top and bottom surface temperature differences Δ T act (Δ T is positive when the top surface temperature is greater than the bottom surface).
According to structural mechanics, the fixed end bending moments corresponding to the three loads all meet the requirement
Figure BDA0002752277880000247
And are each equal to
Figure BDA0002752277880000248
Where α and h are the material linear expansion coefficient and the cross-sectional height, respectively. Mixing the above
Figure BDA0002752277880000249
The corner z of the n-span continuous beam bridge can be obtained by substituting the formula (50) and the formula (51) j And bending moment M j . According to the expressions of the expressions (50) and (51), under the action of uniformly distributed load q and midspan concentrated force F, the node corner z of the continuous beam j Inversely proportional to section bending stiffness EI and node bending moment M j Is irrelevant to EI; and under the action of temperature, the node angle z j EI independent, node bending moment M j Proportional to EI.
While the foregoing is directed to the preferred embodiment of the present invention, it will be appreciated by those skilled in the art that various changes and modifications may be made therein without departing from the principles of the invention as set forth in the appended claims.

Claims (4)

1. A method for calculating influence lines of an equal-span equal-section continuous beam is characterized by comprising the following steps of: the method comprises the following steps:
(1) Determining the solution of the multi-span continuous beam when the leftmost beam end is subjected to the action of concentrated bending moment:
for n-span equal-section continuous beam, the bending moment M is applied to the leftmost end C Angle of rotation z acting at node j j,0 Comprises the following steps:
Figure FDA0003828676360000011
wherein: n is the total span of the continuous beam; j is a node number, and j =0,1, \ 8230;, n; z is a radical of j,0 Is the angle of rotation of the node j, clockwise rotation being positive, 0 in the subscript representing the bending moment M C Acting on node 0; m C The beam end concentrates bending moment, so that the tension of the lower edge of the main beam is positive; i.e. i 0 Is the bending linear stiffness per span of the continuous beam, defined as i 0 ＝EI/l 0 (ii) a Wherein E is the elastic modulus of the main beam material, I is the bending moment of inertia of the main beam section, and l 0 Is the span of each span;
Figure FDA0003828676360000012
wherein constant is
Figure FDA0003828676360000013
Figure FDA0003828676360000014
Figure FDA0003828676360000015
And
Figure FDA0003828676360000016
the superscripts n-j and n of (a) represent the indices;
bending moment M C Moment M acting on the leftmost end of the continuous beam at node j j,0 Comprises the following steps:
Figure FDA0003828676360000017
wherein: m j,0 Is the bending moment at the joint j, so that the tension of the lower edge of the main beam is positive, and 0 in the subscript represents the bending moment M C Acting on node 0;
Figure FDA0003828676360000018
bending moment M C Acting on the leftmost end of the continuous beam at x = (k + xi) l 0 Vertical displacement at position D k+ξ,0 Comprises the following steps:
Figure FDA0003828676360000019
wherein: d k+ξ,0 Is positioned in the k +1 midspan xi l 0 The vertical displacement at the position takes the vertical downward displacement as positive, wherein k =0,1, \8230, n-1,0 ≦ ξ ≦ 1; in the subscripts, 0 represents a bending moment M C Acts on node 0; g is a radical of formula 1 ＝ξ(1-ξ) 2 ，g 2 ＝ξ 2 (1-ξ)；
Figure FDA00038286763600000110
Figure FDA00038286763600000111
Figure FDA00038286763600000112
And
Figure FDA00038286763600000113
the superscripts n-k and n-k-1 of (a) represent indices;
(2) Determining the solution of the multi-span continuous beam when the rightmost beam end is subjected to the action of concentrated bending moment according to the mirror image relationship:
for the n-span equal-section continuous beam, the rightmost end of the beam is subjected to bending moment M C Angle of rotation z acting at node j j,n Bending moment M j,n And when x = (k + ξ) l 0 Vertical displacement at position D k+ξ,n Respectively as follows:
Figure FDA0003828676360000021
Figure FDA0003828676360000022
Figure FDA0003828676360000023
wherein: in the subscripts, n represents a bending moment M C Acting on a node n;
Figure FDA0003828676360000024
Figure FDA0003828676360000025
Figure FDA0003828676360000026
and
Figure FDA0003828676360000027
the superscripts j, k and k +1 of (a) represent the indices;
(3) Determining the beam end rotational stiffness of the multi-span continuous beam:
for the n-span equal-section continuous beam, the bending moment required to be applied when the leftmost end generates a unit corner is defined as the beam end rotating rigidity K n From step (1), K can be obtained n The calculation formula of (c) is:
Figure FDA0003828676360000028
wherein: k n Is the rotational stiffness of the beam end, so that the tension of the lower edge of the main beam is positive,
Figure FDA0003828676360000029
according to the mirror image relationship, the bending moment required to be applied when the unit corner occurs at the rightmost end of the n-span continuous beam is-K n ；
(4) Determining a node bending moment influence line of the multi-span continuous beam according to the mutual equality theorem of work:
for the n-span equal-span uniform-section continuous beam, the bending moment influence line of the n-span equal-section continuous beam at the joint j is marked as M j (x) J =0,1, \8230, n, then for x = (k + xi) l 0 There are the following calculation formulas:
Figure FDA00038286763600000210
(5) Determining analytical formulas of other influence lines according to the node bending moment influence lines in the step (4):
n-span equal-section continuous beam is y = (r + lambda) l 0 The influence of the bending moment at the location is marked M r+λ (x) Where r =0,1, \8230, n-1,0 ≦ λ ≦ 1, and the argument x = (k + ξ) l 0 ，k＝0,1,…,n-1，0≤ξ≤1；M r+λ (x) The expression of (a) is:
Figure FDA0003828676360000031
wherein:
Figure FDA0003828676360000032
Figure FDA0003828676360000033
Figure FDA0003828676360000034
Figure FDA0003828676360000035
and
Figure FDA0003828676360000036
the superscripts n-r, n-r-1, r and r +1 of (a) represent indices;
n span equal section continuous beam is y = (r + lambda) l 0 The shear force influence line at the position is denoted as Q r+λ (x) Setting the shear force to make the beam segment rotate clockwise as positive, then:
Figure FDA0003828676360000037
wherein:
Figure FDA0003828676360000038
shear force influence line Q r+λ (x) At x = (r + λ) l 0 The presence of a mutation;
the reaction influence line of the n-span uniform-section continuous beam support p is denoted as R p (x) Where p =0,1, \8230;, n, with the counter force being positive upwards, then:
when p = 0:
Figure FDA0003828676360000041
wherein,
Figure FDA0003828676360000042
Figure FDA0003828676360000043
and
Figure FDA0003828676360000044
the superscripts n-1 and 1 denote indices;
when p =1,2, \8230;, n-1:
Figure FDA0003828676360000045
wherein:
Figure FDA0003828676360000046
Figure FDA0003828676360000047
Figure FDA0003828676360000048
Figure FDA0003828676360000049
Figure FDA00038286763600000410
Figure FDA00038286763600000411
and
Figure FDA00038286763600000412
the superscripts p-1, p, n-p-1, n-p and p +1 denote indices;
when p = n:
Figure FDA00038286763600000413
n span equal sectionPlane continuous beam at y = (r + lambda) l 0 The influence line of the displacement at the position is marked D r+λ (x) The expression is as follows:
Figure FDA0003828676360000051
wherein f is 1 ＝λ(1-λ)(1+λ)，f 2 λ (1- λ) (2- λ), and:
Figure FDA0003828676360000052
n-span equal-section continuous beam is y = (r + lambda) l 0 The influence line of the rotation angle at the position is represented as theta r+λ (x) The expression is as follows:
Figure FDA0003828676360000053
wherein f is 1 ′＝1-3λ 2 ，f 2 ′＝3λ 2 -6 λ +2, and:
Figure FDA0003828676360000054
the continuous beam is a straight-line continuous beam with equal cross sections and the span and the material are the same, and the shear deformation is not considered in analysis.
2. The method for calculating the influence line of the uniform-span uniform-section continuous beam as claimed in claim 1, wherein the influence line is calculated by the following steps: deducing a corner analytical formula of each node when the n-span equal-span uniform-section continuous beam is under the load action of the u-span according to the beam end rotational stiffness in the step (3):
Figure FDA0003828676360000061
wherein: n is a positive integer, u =1,2, \ 8230;, n; j is a node number, and j =0,1, \ 8230;, n; z is a radical of j Is the corner of node j, and clockwise rotation is positive;
Figure FDA0003828676360000062
and
Figure FDA0003828676360000063
respectively setting the fixed end bending moment of the u-th span main beam under the action of load, and taking the tension of the lower edge of the section as positive;
Figure FDA0003828676360000064
Figure FDA0003828676360000065
Figure FDA0003828676360000066
and
Figure FDA0003828676360000067
the superscripts n-u +1, n-u, u-1 and u denote indices;
according to the equation of angular displacement in structural mechanics, from z j And obtaining the bending moment of each node of the main beam, and further calculating the internal force and displacement of any section position.
3. The method for calculating the influence line of the uniform-span uniform-section continuous beam as claimed in claim 2, wherein the influence line is calculated by the following steps: when the n-span equal-section continuous beam is under the load action on the u-span, and the fixed end bending moment generated by the load meets the requirement
Figure FDA0003828676360000068
When the condition is satisfied, the rotation angle z of each node is j Bending moment M j The analytical formulas of (a) are respectively as follows:
Figure FDA0003828676360000069
Figure FDA00038286763600000610
wherein,
Figure FDA00038286763600000611
4. the method for calculating the influence line of the uniform-span uniform-section continuous beam according to claim 2, characterized in that: when the load of each span of the n-span equal-section continuous beam is the same and the fixed end bending moment generated by the load form meets the requirement
Figure FDA00038286763600000612
When the condition (2) is satisfied, all the node rotation angles z 'after the respective cross-loading effects are considered' j And bending moment M' j The analytical formulas are respectively as follows:
Figure FDA0003828676360000071
Figure FDA0003828676360000072
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