理論力學(xué)C Hapter 5

1、CHAPTER 5Equilibrium Analysis in Two Dimensions(兩維空間的平衡分析)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 (兩投影一矩式)000BAxMMFOne force equation and two moment equations (一投影兩矩式)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 平面拱架由三段不計(jì)自重的剛性桿經(jīng)C、D處鉸接而成，其中CD段與曲桿

12、ABC的BC段同為半徑為R的曲桿。已知：AB = ED = AE = BD =2R = 2m, q q 6060o o, 作用于點(diǎn)B B的水平力為F=2kN, 作用于AB段的均布載荷為q=1kN/m，作用于DE段的力偶之力偶矩為M= 5kNm。試求：固定端約束E處的約束反力。Cq qDBAOEqFM例6 圖示平面結(jié)構(gòu)由丁字形梁ABC、直梁CE與支桿DH組成，C、D點(diǎn)為鉸接，均不計(jì)各桿自重。已知q=200kN/m，P=100kN，M=50kN.m，L=2m。試求固定端A處反力。ABCDEHLLLLLP30qM45例7 圖示平面機(jī)構(gòu)由四根重均為P = 50N，長度均為2L的勻質(zhì)剛桿鉸接而成。線性彈

13、簧原長為L=1m剛性系數(shù)為k =100N/m，B端擱置在光滑的水平面上。求系統(tǒng)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(靜定與靜不定問題)Nu-the number of unknowns Ni -the number of independent equilibrium equationsNiNu NiNu NiNu NiNu EndACHLLaPqBDaqACHLLaPqBDaACHLLaPBDaStatically determinate ?Statically indeterminate ? 5.7 Simple trusses (簡單桁架)錢塘江橋。全長1453 米。中國第一座現(xiàn)代 化公路鐵路兩用雙層 鋼桁架橋梁。 武漢長江大橋。全長 1679米。于

15、1957年建 成?？缍?28米。 英國福斯灣橋。鋼懸 臂桁架雙線鐵路橋。 跨度521米。1890年 建成。 北京首都國際機(jī)場 航空港內(nèi)鋼結(jié)構(gòu)飛 機(jī)庫。 衛(wèi)星發(fā)射塔。法國埃菲爾鐵塔。 ZT120型塔式起重機(jī)焊接焊接高壓線塔高壓線塔鉚接鉚接1.What is a truss ?A truss is a structure composed of slender members joined together at their end point.桁架的實(shí)際節(jié)點(diǎn)理想節(jié)點(diǎn)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 ?(簡單桁架)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(結(jié)點(diǎn)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