东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A05 Analysis of Structures

Analysis of Structures

Analysis of Structures

Contents·Introduction（绪论）·Definition of a Truss（桁架）·SimpleTrusses（简单桁架）·Analysis of TrussesbytheMethodof Joints（由节点法分析桁架）·Zero-forceMembers（零力杆）·AnalysisofTrussesbytheMethodof Sections（由截面法分析桁架）·AnalysisofFrames（刚架分析）2

Contents • Introduction（绪论） • Definition of a Truss（桁架） • Simple Trusses（简单桁架） • Analysis of Trusses by the Method of Joints（由节点法分析桁架） • Zero-force Members（零力杆） • Analysis of Trusses by the Method of Sections（由截面法分析桁架） • Analysis of Frames（刚架分析） 2

LntroductionD: For the equilibrium of structures made of severalEconnected parts, the internal forces as well the externalforces are considered.: In the interaction between connected parts, Newton's 3rdLaw states that the forces of action and reactionbetween bodies in contact have the same magnitude,Esame line of action, and opposite sense.: Three categories of engineering structures are considered:a) Frames: contain at least one one multi-force member,i.e., member acted uponby 3 or more forces.b) Trusses: formed from two-force members, i.e.,straight members with end point connections)Machines:structures containingmovingpartsdesigned to transmit and modify forces.3

Introduction • For the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered. • In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. • Three categories of engineering structures are considered: a) Frames: contain at least one one multi-force member, i.e., member acted upon by 3 or more forces. b) Trusses: formed from two-force members, i.e., straight members with end point connections c) Machines: structures containing moving parts designed to transmit and modify forces. 3

Definition of a Truss. A truss consists of straight members connected atjoints. No member is continuous through a joint.B: Most structures are made of several trusses joinedtogether to form a space framework. Each trusscarries those loads which act in its plane and maybe treated as a two-dimensional structure.万: Bolted or welded connections are assumed to bepinned together. Forces acting at the member endsreduce to a single force and no couple. Only two-force members are considered. When forces tend to pull the member apart, it is intension. When the forces tend to compress themember, it is in compression.4

Definition of a Truss • A truss consists of straight members connected at joints. No member is continuous through a joint. • Bolted or welded connections are assumed to be pinned together. Forces acting at the member ends reduce to a single force and no couple. Only two￾force members are considered. • Most structures are made of several trusses joined together to form a space framework. Each truss carries those loads which act in its plane and may be treated as a two-dimensional structure. • When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression. 4

Simple Trusses. Arigid truss will not collapse underthe application of a load.DA.Asimpletrussis constructedbysuccessively adding two members andone connection to the basic triangularDBtruss. In a simple truss, m =2n -3 wherem isthetotal number of membersand n is the number of jointsBEDF5

Simple Trusses • A rigid truss will not collapse under the application of a load. • A simple truss is constructed by successively adding two members and one connection to the basic triangular truss. • In a simple truss, m = 2n - 3 where m is the total number of members and n is the number of joints. 5

Analysis of Trusses by the Method of Joints: Dismember the truss and create a freebodydiagram for each member and pin. The two forces exerted on each member areequal, have the same line of action, andopposite sense.RB. Forces exerted by a member on the pins orjoints at its ends are directed along the memberand equal and opposite.. Conditions of equilibrium on the pins provide2n equations for 2n unknowns. For a simpletruss.2n=m+3.MaysolveformmemberBforces and 3 reaction forces at the supports.BB.Conditionsforequilibriumforthe entiretrussprovide 3 additional equations which are notindependent of the pin equations.6

Analysis of Trusses by the Method of Joints • Dismember the truss and create a freebody diagram for each member and pin. • The two forces exerted on each member are equal, have the same line of action, and opposite sense. • Forces exerted by a member on the pins or joints at its ends are directed along the member and equal and opposite. • Conditions of equilibrium on the pins provide 2n equations for 2n unknowns. For a simple truss, 2n = m + 3. May solve for m member forces and 3 reaction forces at the supports. • Conditions for equilibrium for the entire truss provide 3 additional equations which are not independent of the pin equations. 6

Zero-force Members.Forces in opposite members intersecting inFAB1two straight lines at a joint are equal.FAC: The forces in two opposite members areFAEequal when a load is aligned with a thirdmember. The third member forceis equalFADCto the load (including zero load)FAC.Theforces intwo members connected at ajoint are equal if the members are alignedand zero otherwise.: Recognition of joints under special loadingconditions simplifies a truss analysis25kN150kNF25kNHREGC20kN7

Zero-force Members • Forces in opposite members intersecting in two straight lines at a joint are equal. • The forces in two opposite members are equal when a load is aligned with a third member. The third member force is equal to the load (including zero load). • The forces in two members connected at a joint are equal if the members are aligned and zero otherwise. • Recognition of joints under special loading conditions simplifies a truss analysis. 7

Sample ProblemSOLUTION:1000lb20001b. Based on a free-body diagram of the-12 ft-12ft-entire truss, solve the 3 equilibriumBACequations for the reactions at E and C8 ft+. Joint A is subjected to only two unknownDEmemberforces.Determinethesefromthejoint equilibrium requirements-12ft6ft6 ft:In succession,determineunknownmember forces at joints D, B, and E fromjoint equilibrium requirements.Using the method of joints, determinethe force in each member of the truss.: All member forces and support reactionsare known at joint C. However, the jointequilibrium requirements may be appliedtochecktheresults.8

Sample Problem Using the method of joints, determine the force in each member of the truss. SOLUTION: • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. • Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. • In succession, determine unknown member forces at joints D, B, and E from joint equilibrium requirements. • All member forces and support reactions are known at joint C. However, the joint equilibrium requirements may be applied to check the results. 8

SOLUTION2000lb1000lb中12ft12ftCBA. Based on a free-body diagram of the entire truss,Csolve the 3 equilibrium equations for the reactions8ft+at E and C.DEEZMc=0-12ft6ft6ft= (2000 1b)(24 ft)+ (1000 1b)(12 ft)- E(6 ft)E=10,0001b个ZFx=0=Cx=0Z F, = 0= -2000 1b-1000 1b+10,0001b +C,C, = 7000 1b I9

SOLUTION: • Based on a free-body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. 2000 lb24 ft 1000 lb12 ft 6 ft 0 E MC      E 10,000 lb Fx  0  Cx Cx  0 Fy     Cy 0 2000 lb -1000 lb 10,000 lb Cy  7000 lb 9

1000lh2000lb-12ft-12ftBAC8ft+DEE12ft6ft6ft: Joint A is subjected to only two unknown2000bmemberforces.Determinethesefromthejoint equilibrium requirements.3FAB2000lb52000 1bFABFADFAB =15001b TFAD3FAD435FABFAD = 25001b CThere are now only two unknown memberFDEFDA=2500lbFDBforces at joint D.FDB554FDB = FDAFDB = 25001b T33FDEFDAFDE = 2)FDAFDE = 30001b C10

• Joint A is subjected to only two unknown member forces. Determine these from the joint equilibrium requirements. 4 3 5 2000 lb FAB FAD   F C F T AD AB 2500 lb 1500 lb   • There are now only two unknown member forces at joint D. DE   DA DB DA F F F F 5 3  2  F C F T DE DB 3000 lb 2500 lb   10