Lecture OF Analysis OF PIN- Jointed Plane Frames PDF

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Lectures notes in Statics of Rigid Bodies ANALYSIS OF PIN-JOINTED PLANE FRAMES PIN-JOINTED PLANE FRAMES A pin-jointed frame is a structure made up of slender member pin connected at end and is capable of taking loads at joints. A frame in which all the members lie in a single plane is called a plane frame. If all members of a frame do not lie in a single plane, the frame is called space frame. Assumptions made in the analysis of pin jointed frame are: 1. The ends of the members are pin connected (hinged-ends). 2. The loads act only at the joints. 3. Self-weight of the members is negligible. 4. Cross-sections of all members are uniform. Methods OF FORCE ANALYSIS Two methods for the force analysis of simple trusses will be given. The external reactions are usually determined first, by applying the equilibrium equations to the truss as a whole. Then the force analysis of the remainder of the truss is performed. 

Method of Joints

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Method of Sections

Method of joints This method for finding the forces in the members of a truss consists of satisfying the conditions of equilibrium for the forces acting in the connecting pin of each joint. The method therefore deals with the equilibrium of concurrent forces, and only two independent equilibrium equations are involved.

Examples: 1. Determine the force in each member of the simple equilateral truss shown. Answer: AB = 736 N (T), AC = 368 N (T), BC = 736 N (C)

2. Determine the force in each member of the loaded truss. Answer: AB = DE = 96 kN (C), AH = EF = 75 kN (T), BC = CD = 75 kN (C), BH = CG = DF = 60 kN (T), CH = CF = 48 kN (C), GH = FG = 112.5 kN (T)

3. Determine the magnitude and nature of the force in the members BC, GC, and GF of the inclined truss shown. Answer: BC = 24 kN (C), GC = 27.7 kN (C), GF = 20.8 kN (T)

4. A king post truss of 8 m span is loaded as shown in the figure. Find the forces in the members BF and DH. Answer: BF = DH = 0 kN

5. Determine the force in each member of the loaded truss. Answer: AB = 12 kN(T), AE = 3 kN (C), BC = 5.20 kN (T), BD = 6 kN (T), BE = 5.20 kN (C), CD = DE = 6 kN (C)

ZERO-FORCE MEMBERS Truss analysis using the method of joints is greatly simplified if we can first identify those members which support no loading. These zero-force members are used to increase the stability of the truss during construction and to provide added support if the loading is changed. The zero-force members of a truss can generally be found by inspection. Rule 1: If only two members form a truss joint and no external load or support reaction is applied to the joint, the two members must be zero-force members. Rule 2: If three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied to the joint.

Method of sections When analyzing plane trusses by the method of joints, we need only two of the three equilibrium equations because the procedures involve concurrent forces in each joint. We can take advantage of the third or the moment equation of equilibrium by selecting an entire section of the truss for the free body in equilibrium under the action of a non-concurrent system of forces. This method of sections has the basic advantage that the force in almost any desired member may be found directly from an analysis of a section which has cut that member. Thus, it is not necessary to proceed with the calculation from joint to joint until the member in question has been reached. In choosing a section of the truss, we note that, in general, not more than three members whose forces are unknown should be cut, since there are only three available independent equilibrium equations.

Examples: 1. The figure shows a Warren girder consisting of seven members each of 3 m length freely supported at its endpoint. The girder is loaded at B and C shown. Find the forces in all the members of the girder, indicating whether the force is compressive or tensile. Answer: AB = 2.89 kN (C), AE = 1.44 kN (T), CD = 4.04 kN (C), DE = 2.02 kN (T), BE = 0.58 kN (T), BC = 1.73 kN (C), CE = 0.58 kN (C) 2. Determine the nature and magnitude of the forces in members 1, 2, and 3. Answer: P1 = 625 N (C), P2 = 750 N (T), P3 = 500 N (T)

3. Determine the force in member CG. Answer: 14.14 kips (T)...