理论力学C Hapter 5

1、CHAPTER 5Equilibrium Analysis in Two Dimensions(两维空间的平衡分析)5.1 Introduction 1. Analysis of single bodies. (only one FBD is employed and 3 independent equilibrium equations are used in problem solving process ) 2. Analysis of composite bodies. (frames and machines, more than one FBD and at least 3 equ

2、ations ) 3. Analysis of plane trusses. ( Joint method, in fact concurrent force system. Section method , coplanar force system. )5.2 Analysis of single bodiesSample 1 Beam AB as shown in Fig. Find the reactions at support A and B.0 xF060cos5 . 1AxF75. 0AxF0)F(AM0)5 . 15 . 2(60sin5 . 15 . 122 . 15 .

3、2BF75. 3)460sin5 . 132 . 1 (5 . 21BF0yF060sin5 . 12BAyFF45. 075. 360sin5 . 12AyFSolution FBD of beam Equilibrium equationsSample 2 given kN240Pcm100acm140bcm140dcm100e55Solution FBD of beam Equilibrium equations0 xF0sinPF0yF0cosPFFNBNA0)F(AM0sincos)(ePaPbaFFdNBkNF6 .196kN2 .904 . 2/ ) 155sin240155co

4、s2404 . 16 .196()/()sincos(baePaPFdFNBkNaPFFNBNA6 .4755cos2402 .90cosEndFind tension F, and forces supporting wheel A and B. Sample 3 given l,r,P,q=45o, find the reactions at support A and B.Solution FBD of the structure (member BC is a two-force body)0)(FAM022lPlFlTB0 xF0cosPFFBAxq0yF0sinqBAyFPFFro

5、m the equations above we get0)(2PPFBPFAxPFAyEquilibrium equationsEnd5.3 Discussion of Conditions of Equilibrium and The Equilibrium equations in Two Dimension000AyxMFF00ORMFTwo force equations and one moment equation (两投影一矩式)000BAxMMFOne force equation and two moment equations (一投影两矩式)Three moment e

6、quations (叁矩式)000CBAMMM(AB and x are not vertical to each other.)(A、B and C are not collinear.)Sample 4 given a,b,c,P,Q. determine the reactions at bearing A and B.Solution FBD of beam Method 1 Equilibrium equationsBANM00cQbPaNBaQcPbNBBAxxNNF0QPNFAyy0Method 2 Equilibrium equations00QPNFAyy00QcPbaNMB

7、A00QcPbaNMAxBEndSample 5 The four-bar mechanism as shown in the Fig. given directions of P and Q . Determine the relationship between P and Q when the mechanism is in equilibrium. 0EMAEQBEPo30cosBEAE2PQ23261. 0322QPSolution1 FBD of the whole body.Equilibrium equations EndSolution2 FBDs of joint A an

8、d B5.4 Reduction of distributed forcesmqlxq lmRlqdxqF021lRxdxqhF0lh32End5.5 Analysis of composite bodiesSample 1 given F=20kN, q=10kN/m, M=20kN.m, L=1m ,all weights negligible. Find the reactions at support A and B.Solution FBDs of the right part and the whole body.AMqCLLLLB60DF30。AMqCB60DF30。MAFAxF

9、AyFBqCB60DF30。FcxFcyAccording the FBD of right part write the moment equationFB=45.77kNAccording the FBD of whole body write the equationsFAX=32.89kN, FAY=-2.32kNMA=10.37kN.m 0AM0230cos260sin2 LFqLLFB0430cos360sin22 LFLFLqLMMBA030sin60cosFFFBAx030cos260sinFqLFFBAy 0AM0 xF0yFEndFBqCB60DF30。FcxFcyAMqC

10、B60DF30。MAFAxFAySample2 given l, q, M weights negligible. Determine the reactions at support E.(a)(b)(c)Solution FBDs of the whole structure and the left part.From Fig. (c), write equation0)(FCM0212qllFMBqllMFB21From Fig. (b), write equation0)(FEM0)3(212EBMlqlFM 0 xF0ExF 0yF03EyBFqlFEnd24qlMEqllMFEy

11、5 . 2Solution consider the whole frame FBD shown as Fig (b)Sample3 given a, P, Q. weights negligible. Determine the reactions at support A and B.(a)(b)(c)0212120aPaQaYMBA)(41QPYBQPYMAB4143000QXXFBAx)(410QPXMAC)3(41PQXBFrom the FBD of left part shown as Fig (c)End 例4 平面拱架由三段不计自重的刚性杆经C、D处铰接而成，其中CD段与曲杆

12、ABC的BC段同为半径为R的曲杆。已知：AB = ED = AE = BD =2R = 2m, q q 6060o o, 作用于点B B的水平力为F=2kN, 作用于AB段的均布载荷为q=1kN/m，作用于DE段的力偶之力偶矩为M= 5kNm。试求：固定端约束E处的约束反力。Cq qDBAOEqFM例6 图示平面结构由丁字形梁ABC、直梁CE与支杆DH组成，C、D点为铰接，均不计各杆自重。已知q=200kN/m，P=100kN，M=50kN.m，L=2m。试求固定端A处反力。ABCDEHLLLLLP30qM45例7 图示平面机构由四根重均为P = 50N，长度均为2L的匀质刚杆铰接而成。线性弹

13、簧原长为L=1m刚性系数为k =100N/m，B端搁置在光滑的水平面上。求系统的平衡位置q q。EBAJQHCDkq qExample8 The smooth disk shown in Fig. is pinned at D and has a weight of 20N. Neglecting the weights of the other members, determine the horizontal and vertical components of reaction at pins B and D.5.6 Statically Determinate and Statical

14、ly Indeterminate Problems(静定与静不定问题)Nu-the number of unknowns Ni -the number of independent equilibrium equationsNiNu NiNu NiNu NiNu EndACHLLaPqBDaqACHLLaPqBDaACHLLaPBDaStatically determinate ?Statically indeterminate ? 5.7 Simple trusses (简单桁架)钱塘江桥。全长1453 米。中国第一座现代 化公路铁路两用双层 钢桁架桥梁。 武汉长江大桥。全长 1679米。于

15、1957年建 成。跨度128米。 英国福斯湾桥。钢悬 臂桁架双线铁路桥。 跨度521米。1890年 建成。 北京首都国际机场 航空港内钢结构飞 机库。 卫星发射塔。法国埃菲尔铁塔。 ZT120型塔式起重机焊接焊接高压线塔高压线塔铆接铆接1.What is a truss ?A truss is a structure composed of slender members joined together at their end point.桁架的实际节点理想节点Assumptions for Ideal trusses(1) The members are joined together b

16、y smooth pins.(2) All loadings are applied at the joints.(3) All weights of the member are neglected.2.What is a Ideal truss ?(理想桁架)3.What is a Simple truss ?(简单桁架)A simple truss is constructed by starting with a basic triangle element and connecting two members to form an additional element. So on

17、and so sort.4.The Method of Joints(结点法)5.The Method of Sections(截面法)Example Determine the force in member EB of the roof truss shown in the Fig. indicate whether the member is in tension or compression.NFFMEDEDB30000430sin4400023000410000NFFFFFEBEByEFx20000100030sin300020030cos300030cos0Example determine the force in member EB of the roof truss shown in Fig. Indicate whether the member is in tension or compression. summary Drawing FBDs of the whole body or