CN112307540A - 等截面刚架受等距分布且对称力下关键点弯矩值推导方法 - Google Patents

等截面刚架受等距分布且对称力下关键点弯矩值推导方法 Download PDF

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CN112307540A
CN112307540A CN202011152911.8A CN202011152911A CN112307540A CN 112307540 A CN112307540 A CN 112307540A CN 202011152911 A CN202011152911 A CN 202011152911A CN 112307540 A CN112307540 A CN 112307540A
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何舜
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Abstract

本发明提供了等截面刚架受等距分布且对称力下关键点弯矩值推导方法,该等截面刚架受等距分布且对称力下关键点弯矩值推导方法包括以下步骤:步骤一,先通过查阅我国土木界常用手册,找出单层框架上受单个集中力在任意位置作用(图2)及均布荷载作用(图3)的公式,本发明针对等截面门式刚架在最常见受力状态下,提出关键点内力表达式的推导方法,得出的公式非常简单实用,并填补了力学手册的空白,在无电算条件时用手算能很快求得结果,有较大实用价值和理论价值,且通过延续推导得出本文公式的广泛适用性。

Description

等截面刚架受等距分布且对称力下关键点弯矩值推导方法
技术领域
本发明涉及单层等截面刚架受力技术领域,特别涉及等截面刚架受等距分布且对称力下关键点弯矩值推导方法。
背景技术
在工程设计时间中,经常遇到图1的受力模式——单层框架梁上受等距离分布且对称的集中力(如次梁)的作用,(如钢筋混凝土框架结构的风雨操场、车间等)需要求出各关键点(A、B、C、D)处的弯矩值,以便画出框架的弯矩图,但目前我国土木界常用手册仍是建筑工业出版社的【1】或【2】或【3】(以下简称《手册》),却找不到此受力模式的有关公式,查阅其他类型手册也找不到,所以图1情况受力模式表达式是个空白,《手册》只是单层框架上受单个集中力在任意位置作用图2及均布荷载作用图3的公式,以往结构设计人员用设计手算法计算图1情况受力时,惯用手法是用图2叠加或将集中荷载化成均布荷载用图3情形公式计算,这种方法比较费时,为此我们通过实验发现,其实,通过严密的推导,可以求出图1情况下各关键点的弯矩力学表达式,而且式子很简洁,很实用,并沟通了图1、2和3三者的关系,具体方法是用图2的结果来推导图1,再用图1的结果来推导图3的结果,并且还可以通过求极限来推导单跨梁受力情形图4和图5的受力结果。
发明内容
本发明的主要目的在于提供等截面刚架受等距分布且对称力下关键点弯矩值推导方法,可以有效解决背景技术中的问题。
为实现上述目的,本发明采取的技术方案为:
等截面刚架受等距分布且对称力下关键点弯矩值推导方法,该等截面刚架受等距分布且对称力下关键点弯矩值推导方法包括以下步骤:
步骤一,先通过查阅我国土木界常用手册,找出单层框架上受单个集中力在任意位置作用(图2)及均布荷载作用(图3)的公式;
步骤二,用建筑工程中常用的位移法求出图2的情形,得出如下公式:
Figure BDA0002741712050000021
Figure BDA0002741712050000022
Figure BDA0002741712050000023
Figure BDA0002741712050000024
Figure BDA0002741712050000025
Figure BDA0002741712050000026
(μ为梁柱线刚度之比),
Figure BDA0002741712050000027
上述公式可缩写为:
MC={0.5(v-u)K2+K1}puvL
MD={K1-0.5(v-u)K2}puvL
Figure BDA0002741712050000028
Figure BDA0002741712050000029
Figure BDA00027417120500000210
其中式中:MA代表关键点A处弯矩值,MB代表关键点B处弯矩值,MC代表关键点C处弯矩值,MD代表D处弯矩值,I1代表梁截面惯性矩,I2柱截面惯性矩,u+v=1,H为水平反力,u代表柱刚度;
步骤三,设n为集中荷载把横梁划分成的段数,用步骤二中得出的图(2)的结果来叠加求出N=1,2,…,n的结果:
(1)当n=1时,如上图(6),u1=0,v1=1;u2=1,v2=0;所以Mc=MD=MA=MB=H=0
(2)当n=2时,如上图(7),u3=v3=0.5;
Figure BDA0002741712050000031
所以
Figure BDA0002741712050000032
Figure BDA0002741712050000033
(3)当n=3时,如上图(8),
Figure BDA0002741712050000034
所以
Figure BDA0002741712050000035
Figure BDA0002741712050000036
Figure BDA0002741712050000037
(4)当n=4时,如图(9),利用n=2的结果叠加
Figure BDA0002741712050000038
Figure BDA0002741712050000039
Figure BDA00027417120500000310
Figure BDA00027417120500000311
(5)当n=5时,如图(10),u分别为
Figure BDA00027417120500000312
v分别为
Figure BDA00027417120500000313
分别叠加有
Figure BDA00027417120500000314
Figure BDA00027417120500000315
(6)当n=6时,同理可得:
Figure BDA00027417120500000316
Figure BDA00027417120500000317
(7)当n=7时,同理可得:
Figure BDA0002741712050000041
Figure BDA0002741712050000042
(8)当n=8时,同理可得:
Figure BDA0002741712050000045
Figure BDA0002741712050000043
步骤四,从步骤三中n=1至n=8结果可归纳出(n+1)个p作用下的弯矩及水平反力H有如下规律:
以MC为例:
Figure BDA0002741712050000044
显然,该系数的分母为n2,而分子为0,1,4,10,20,35,56,84….N;
根据以上数据,归纳出分子N的规律如下:
n=1时,N=0=0
n=2时,N=1=0+1
n=3时,N=4=0+1+3
n=4时,N=10=0+1+3+6
n=5时,N=20=0+1+3+6+10
n=6时,N=35=0+1+3+6+10+15
n=7时,N=56=0+1+3+6+10+15+21
n=8时,N=84=0+1+3+6+10+15+21+28
即得出N=0+1+3+6+.....+n(n-1)/2,最后得出下列公式:
Figure BDA00027417120500000510
Figure BDA00027417120500000511
Figure BDA00027417120500000512
所以系数为
Figure BDA0002741712050000051
Figure BDA0002741712050000052
式中(A-1)为标记号;
步骤五,根据步骤一至步骤四的推导,最终得出如下结果:
当有(b+1)个集中力P等距离且对称作用在刚架梁上,如图(1)情形,则有:
Figure BDA0002741712050000053
μ为梁柱线刚度之比。
Figure BDA0002741712050000054
水平反力
Figure BDA0002741712050000055
式中(A-2)为标记号。
优选的,可利用步骤四所得出的(A-1)推导出图3中MC和MD的数值,公式如下:
当n数值很大时,集中荷载趋向于等效均布荷载,即可以令np=qL,得
Figure BDA0002741712050000056
Figure BDA0002741712050000057
代入式(A-1),有
Figure BDA0002741712050000058
当n→∞时,即为均布荷载的情形:
Figure BDA0002741712050000059
(《手册》公式)
这两式就变成了图(3)情形,得出单层刚架在均布荷载作用下MC、MD、及H的表达式;
优选的,当取μ=0,即柱刚度无限大,则刚架图1变成了两端固端的单跨梁,如图4所示,则如下公式:
Figure BDA0002741712050000061
Figure BDA0002741712050000062
代入式(A-1)
Figure BDA0002741712050000063
(《手册》公式)
上式即为两端固端单跨梁受(n+1)个等距离集中力P作用时的支座弯矩。
优选的,
令qL=np,有
Figure BDA0002741712050000064
(当n数值很大时,集中荷载趋向于等效均布荷载,如图(5)示)
Figure BDA0002741712050000065
代入上式,则
Figure BDA0002741712050000066
当n→∞式,
Figure BDA0002741712050000067
(《手册》公式)
与现有技术相比,本发明具有如下有益效果:
单层等截面框架梁受等距离分布且对称的集中力的作用下,各关键点的力学表达式很简单,比起受单个集中力在任意位置情形还要简单得多,即使是计算机相当普及的今天,在计算这种情形时用计算机也不如代入公式用手算来得快,所以公式A-1、公式A-2和公式A-3有相当大的实用价值,在理论上,通过求极限的方式分别推导出单层框架受均布荷载图3及单跨梁分别在等距离集中力图4和均布荷载图5作用下三种情形的力学表达式,说明图3、图4、图5分别只是图1情形的其中一种特例,从而也证明了本文所得出结果的广泛代表性,适用性。
附图说明
图1为本发明等截面刚架受等距分布且对称力下关键点弯矩值推导方法的示意图;
图2为单层等截面刚架受受单个集中力在任意位置作用下的示意图;
图3为单层等截面刚架均布荷载作用下示意图;
图4为两端为固端单跨梁刚架受等距分布且对称力作用下示意图;
图5为两端为固定单跨梁刚架均布荷载作用下示意图;
图6为单层等截面刚架上整段横梁受集中荷载作用下示意图;
图7为单层等截面刚架受集中荷载将刚架横梁分两段示意图;
图8为单层等截面刚架受集中荷载将刚架横梁分三段示意图;
图9为单层等截面刚架受集中荷载将刚架横梁分四段示意图;
图10为单层等截面刚架受集中荷载将刚架横梁分五段示意图。
具体实施方式
为使本发明实现的技术手段、创作特征、达成目的与功效易于明白了解,下面结合具体实施方式,进一步阐述本发明。
实施例1
本实施利涉及等截面刚架受等距分布且对称力下关键点弯矩值推导方法,该等截面刚架受等距分布且对称力下关键点弯矩值推导方法包括以下步骤:骤一,先通过查阅我国土木界常用手册,找出单层框架上受单个集中力在任意位置作用(图2)及均布荷载作用(图3)的公式;
步骤二,用建筑工程中常用的位移法求出图2的情形,得出如下公式:
Figure BDA0002741712050000071
Figure BDA0002741712050000072
Figure BDA0002741712050000073
Figure BDA0002741712050000074
Figure BDA0002741712050000075
Figure BDA0002741712050000076
(μ为梁柱线刚度之比),
Figure BDA0002741712050000077
上述公式可缩写为:
MC={0.5(v-u)K2+K1}puvL
MD={K1-0.5(v-u)K2}puvL
Figure BDA0002741712050000081
Figure BDA0002741712050000082
Figure BDA0002741712050000083
其中式中:MA代表关键点A处弯矩值,MB代表关键点B处弯矩值,MC代表关键点C处弯矩值,MD代表D处弯矩值,I1代表梁截面惯性矩,I2柱截面惯性矩,u+v=1,H为水平反力,u代表柱刚度;
步骤三,设n为集中荷载把横梁划分成的段数,用步骤二中得出的图(2)的结果来叠加求出N=1,2,…,n的结果:
(1)当n=1时,如上图(6),u1=0,v1=1;u2=1,v2=0;所以Mc=MD=MA=MB=H=0
(2)当n=2时,如上图(7),u3=v3=0.5;
Figure BDA0002741712050000091
所以
Figure BDA0002741712050000092
Figure BDA0002741712050000093
(3)当n=3时,如上图(8),
Figure BDA0002741712050000094
所以
Figure BDA0002741712050000095
Figure BDA0002741712050000096
Figure BDA0002741712050000097
(4)当n=4时,如图(9),利用n=2的结果叠加
Figure BDA0002741712050000098
Figure BDA0002741712050000099
Figure BDA00027417120500000910
Figure BDA00027417120500000911
(5)当n=5时,如图(10).u分别为
Figure BDA00027417120500000912
v分别为
Figure BDA00027417120500000913
分别叠加有
Figure BDA00027417120500000914
Figure BDA00027417120500000915
(6)当n=6时,同理可得:
Figure BDA00027417120500000916
Figure BDA00027417120500000917
(7)当n=7时,同理可得:
Figure BDA0002741712050000101
Figure BDA0002741712050000102
(8)当n=8时,同理可得:
Figure BDA0002741712050000103
Figure BDA0002741712050000104
步骤四,从步骤三中n=1至n=8结果可归纳出(n+1)个p作用下的弯矩及水平反力H有如下规律:
以MC为例:
Figure BDA0002741712050000105
显然,该系数的分母为n2,而分子为0,1,4,10,20,35,56,84….N;
根据以上数据,归纳出分子N的规律如下:
n=1时,N=0=0
n=2时,N=1=0+1
n=3时,N=4=0+1+3
n=4时,N=10=0+1+3+6
n=5时,N=20=0+1+3+6+10
n=6时,N=35=0+1+3+6+10+15
n=7时,N=56=0+1+3+6+10+15+21
n=8时,N=84=0+1+3+6+10+15+21+28
即得出N=0+1+3+6+.....+n(n-1)/2,最后得出下列公式:
Figure BDA0002741712050000111
所以系数为
Figure BDA0002741712050000112
Figure BDA0002741712050000113
式中(A-1)为标记号;
步骤五,根据步骤一至步骤四的推导,最终得出如下结果:
当有(n+1)个集中力P等距离且对称作用在刚架梁上,如图(1)情形,则有:
Figure BDA0002741712050000114
μ为梁柱线刚度之比。
Figure BDA0002741712050000115
水平反力
Figure BDA0002741712050000116
式中(A-2)为标记号;
还可以利用步骤四所得出的(A-1)推导出图3中MC和MD的数值,公式如下:
当n数值很大时,集中荷载趋向于等效均布荷载,即可以令np=qL,得
Figure BDA0002741712050000117
Figure BDA0002741712050000118
代入式(A-1),有
Figure BDA0002741712050000119
当n→∞时,即为均布荷载的情形:
Figure BDA00027417120500001110
(《手册》公式)
;以上得出的公式就变成了图(3)情形,得出单层刚架在均布荷载作用下MC、MD、及H的表达式;当取μ=0,即柱刚度无限大,则刚架图1变成了两端固端的单跨梁,如图4所示,则如下公式:
Figure BDA0002741712050000121
Figure BDA0002741712050000122
代入式(A-1)
Figure BDA0002741712050000123
(《手册》公式)
上式即为两端固端单跨梁受(n+1)个等距离集中力P作用时的支座弯矩。
令qL=np,有
Figure BDA0002741712050000124
(当n数值很大时,集中荷载趋向于等效均布荷载,如图(5)示)
Figure BDA0002741712050000125
代入上式,则
Figure BDA0002741712050000126
当n→∞式,
Figure BDA0002741712050000127
(《手册》公式);
根据以上实施例中所得出的结果:单层等截面框架梁受等距离分布且对称的集中力的作用下,各关键点的力学表达式很简单,比起受单个集中力在任意位置情形还要简单得多,即使是计算机相当普及的今天,在计算这种情形时用计算机也不如代入公式用手算来得快,所以公式A-1、公式A-2和公式A-3有相当大的实用价值,在理论上,通过求极限的方式分别推导出单层框架受均布荷载图3及单跨梁分别在等距离集中力图4和均布荷载图5作用下三种情形的力学表达式,说明图3、图4、图5分别只是图1情形的其中一种特例,从而也证明了本文所得出结果的广泛代表性,适用性。
以上显示和描述了本发明的基本原理和主要特征和本发明的优点。本行业的技术人员应该了解,本发明不受上述实施例的限制,上述实施例和说明书中描述的只是说明本发明的原理,在不脱离本发明精神和范围的前提下,本发明还会有各种变化和改进,这些变化和改进都落入要求保护的本发明范围内。本发明要求保护范围由所附的权利要求书及其等效物界定。

Claims (4)

1.等截面刚架受等距分布且对称力下关键点弯矩值推导方法,其特征在于,包括以下步骤:
步骤一,先通过查阅我国土木界常用手册,找出单层框架上受单个集中力在任意位置作用(图2)及均布荷载作用(图3)的公式;
步骤二,用建筑工程中常用的位移法求出图2的情形,得出如下公式:
Figure FDA0002741712040000011
Figure FDA0002741712040000012
Figure FDA0002741712040000013
Figure FDA0002741712040000014
Figure FDA0002741712040000015
Figure FDA0002741712040000016
(μ为梁柱线刚度之比),
Figure FDA0002741712040000017
上述公式可缩写为:
MC={0.5(v-u)K2+K1}puvL
MD={K1-0.5(v-u)K2}puvL
Figure FDA0002741712040000018
Figure FDA0002741712040000019
Figure FDA00027417120400000110
其中式中:MA代表关键点A处弯矩值,MB代表关键点B处弯矩值,MC代表关键点C处弯矩值,MD代表D处弯矩值,I1代表梁截面惯性矩,I2柱截面惯性矩,u+v=1,H为水平反力,u代表柱刚度;
步骤三,设n为集中荷载把横梁划分成的段数,用步骤二中得出的图(2)的结果来叠加求出N=1,2,…,n的结果:
(1)当n=1时,如上图(6),u1=0,v1=1;u2=1,v2=0;所以MC=MD=MA=MB=H=0
(2)当n=2时,如上图(7),u3=v3=0.5;
Figure FDA0002741712040000021
所以
Figure FDA0002741712040000022
Figure FDA0002741712040000023
(3)当n=3时,如上图(8),
Figure FDA0002741712040000024
所以
Figure FDA0002741712040000025
Figure FDA0002741712040000026
Figure FDA0002741712040000027
(4)当n=4时,如图(9),利用n=2的结果叠加
Figure FDA0002741712040000028
Figure FDA0002741712040000029
Figure FDA00027417120400000210
Figure FDA00027417120400000211
(5)当n=5时,如图(10),u分别为
Figure FDA00027417120400000212
v分别为
Figure FDA00027417120400000213
分别叠加有
Figure FDA00027417120400000214
Figure FDA00027417120400000215
(6)当n=6时,同理可得:
Figure FDA0002741712040000031
Figure FDA0002741712040000032
(7)当n=7时,同理可得:
Figure FDA0002741712040000033
Figure FDA0002741712040000034
(8)当n=8时,同理可得:
Figure FDA0002741712040000035
Figure FDA0002741712040000036
步骤四,从步骤三中n=1至n=8结果可归纳出(n+1)个p作用下的弯矩及水平反力H有如下规律:
以MC为例:
Figure FDA0002741712040000037
显然,该系数的分母为n2,而分子为0,1,4,10,20,35,56,84....N;
根据以上数据,归纳出分子N的规律如下:
n=1时,N=0=0
n=2时,N=1=0+1
n=3时,N=4=0+1+3
n=4时,N=10=0+1+3+6
n=5时,N=20=0+1+3+6+10
n=6时,N=35=0+1+3+6+10+15
n=7时,N=56=0+1+3+6+10+15+21
n=8时,N=84=0+1+3+6+10+15+21+28
即得出N=0+1+3+6+.....+n(n-1)/2,最后得出下列公式:
Figure FDA0002741712040000041
所以系数为
Figure FDA0002741712040000042
Figure FDA0002741712040000043
式中(A-1)为标记号;
步骤五,根据步骤一至步骤四的推导,最终得出如下结果:
当有(n+1)个集中力P等距离且对称作用在刚架梁上,如图(1)情形,则有:
Figure FDA0002741712040000044
μ为梁柱线刚度之比。
Figure FDA0002741712040000045
水平反力
Figure FDA0002741712040000046
式中(A-2)为标记号。
2.根据权利要求书1所述的等截面刚架受等距分布且对称力下关键点弯矩值推导方法,其特征在于:可利用步骤四所得出的(A-1)推导出图3中MC和MD的数值,公式如下:
当n数值很大时,集中荷载趋向于等效均布荷载,即可以令np=qL,得
Figure FDA0002741712040000047
Figure FDA0002741712040000048
代入式(A-1),有
Figure FDA0002741712040000049
当n→∞时,即为均布荷载的情形:
Figure FDA00027417120400000410
这两式就变成了图(3)情形,得出单层刚架在均布荷载作用下MC、MD、及H的表达式。
3.根据权利要求书2所述的等截面刚架受等距分布且对称力下关键点弯矩值推导方法,其特征在于:当取μ=0,即柱刚度无限大,则刚架图1变成了两端固端的单跨梁,如图4所示,则如下公式:
Figure FDA0002741712040000051
Figure FDA0002741712040000052
代入式(A-1)
Figure FDA0002741712040000053
上式即为两端固端单跨梁受(n+1)个等距离集中力P作用时的支座弯矩。
4.根据权利要求书2所述的等截面刚架受等距分布且对称力下关键点弯矩值推导方法,其特征在于:
令qL=np,有
Figure FDA0002741712040000054
(当n数值很大时,集中荷载趋向于等效均布荷载,如图(5)示)
Figure FDA0002741712040000055
代入上式,则
Figure FDA0002741712040000056
当n→∞式,
Figure FDA0002741712040000057
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