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Structural Analysis and Bolting Techniques for Trusses

Learn about simple trusses, bridge trusses, and the method of joints in structural analysis using bolting. Understand zero-force members and the method of sections for truss evaluation.

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Structural Analysis and Bolting Techniques for Trusses

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  1. CHAP6. Structural Analysis

  2. Bolting 6.1 Simple Trusses 1. Definition of Truss A structure composed of slender members jointed together at their end points by bolting, welding or pinning. 2. Planar Trusses planar trusses lie in a single and one often need to support roofs and bridges. Bridge truss Roof truss

  3. 3. Assumption for Truss T T Tensile force T C C Compression force (1) all loading are applied at the joints a. members’ weights are neglected (W << external force) b. members’ weights are included (2) members are jointed together by smooth pin (No friction forces) Because of the two assumptions, the truss member is a two-force member T Note: Compression member must be made thicken than tension member because of the buckling or column effect in compression member.

  4. E P C D B A C A D unstable truss 4. Simple Truss A simple truss is constructed by starting with a basic triangular element and connecting two members to form an additional element. Simple form of rigid or stable truss

  5. 6.2 the Method of Joints 1. Objective Evaluate the force acting on each member of a truss structure to analyze or design a truss 2. Concept A truss is in equilibrium, so each of its joint also is in equilibrium Force system acting at pin or joint is coplanar and concurrent • Therefore, we have • member equilibrium is automatically satisfied at each joint • Only force equilibrium ΣFx=0 and ΣFy=0 is necessary

  6. 3. Analysis Procedure Support A is pin constraint. Support C is roller constraint. (1) Draw the free-body diagram of a joint having at least one known force and at most two unknown forces. For the given truss, only joint B is satisfied. F.B.D of joint B

  7. (2) Establish the sense of the unknown forces (a) assume the unknown member forces to be in tension. (b) Determine correct sense of unknown member forces by inspection

  8. (3) Apply the force equilibrium equations ΣFx=0 500+FBC sin45°=0 ΣFy=0 -FAB-FBCcos45°=0 (4) Solve the unknown forces and check their correct sense FBC=- 500/sin45°=-500√2N (in compression) FAB=-FBCcos45°=500N (in tension) (5) Continue to analyze then joints by repeating steps (1) to (4)

  9. choose joint C first, then joint A F.B.D of joint C equilibrium equation at joint C ΣFx=0 500√2 cos45°-FAC=0 ΣFy=0 Cy-500√2 sin45°=0 FAC=500N (in tension);Cy=500N F.B.D joint A Ax=Ay=500N

  10. D F E C A B 6-3 zero-force members 1. zero-force member A member supports no loading, which is used to increase the stability of the truss and to provide support if the applied loading is changed. 2. Rules for determining zero-force members (1) If only two members form a truss joint and no external load or support reaction is applied to the joint. Ex. Joint A and joint D areformed by two members AF &AB, ED&CD respectively and no loading is applied. Hence, members AF, A, ED and CD are zero-force members.

  11. F E b C B P F.B.D of joint A Equations of equilibrium FAF • ΣFx=0 , so FAB=0 • ΣFy=0 , so FAF=0 FAB A Equations of equilibrium F.BD of joint D • ΣFx=0, FED+FDC cosθ=0 • ΣFy=0, FDC sinθ=0 • So, FDC=0,FED=0 D FED x FDC y The equivalent system is given as shown.

  12. P E D C A B (2) If three members form a truss joint for which two of the members are collinear , the third member is zero-force members provided no external force or support reaction is applied to the joint. EX: Joint D :3 members AD, ED and CD (ED &DC collinear) Joint C :3 members AD, DC and CB (DC& CB collinear) No external force or support reaction is applied to the joint D & C. ∴AD and AC zero-force members

  13. 6-4 The method of section T In equilibrium F=T F Internal tension force F In equilibrium T F=T In equilibrium T 1. Purpose Determine the loadings acting within a body . 2. Principle If a body is equilibrium , then any part of the body is also in equilibrium. T Imaginary section Apply the equations of equilibrium to the sectioned part to determine the loading at the section.

  14. 500N B 2m A 45 C 2m B 500N y 2m x A C Ax Ay Cy 3. Analysis procedure (1) Determined the external reactions at the constraints F.B.D of entire truss. F.B.D. of entire truss Equations of equilibrium 500-Ax=0 + Cy-Ax=0 + 500x2-Cyx2=0 +

  15. B 500N (1) 500N A (2) C 500N (3) 500N 500N C 500N A 500N 500N (2) Cut or section the truss through not more than three members in which the forces are unknown. (Because three independent equilibrium equations for three unknowns.) three possible ways of section for given truss (3) Draw the free-body diagram of the part of sectioned truss which has the least number of forces acting on it. Section(1)

  16. + 500N-FBCsin45=0 + FBCcos45-FAB=0 (in compression) FBC=500 (in tension) FAB=500N (4) Establish the sense of the unknown member force 500N y 45 x F FBC AB (5) Apply the equations of equilibrium and check the correct sense of solve forces. • Moments should be summed about a point lying at the intersection of the lines of actions of two unknown forces. The third unknown is determined directly from the moment equation. • Forces may be summed perpendicular to the direction of the two unknown forces which are parallel.

  17. B C A y x FAC 500N + + (6) Continue to analyze other new sections by repeating steps(1)to(5). Section (2) F.B.D. of right part of section (2) Equations of equilibrium So, FAC=500N

  18. 6.6 Frames and Machines 1. Definition (1)Frames A stationary structure composed of pin-connected multiforce members is used to support loads. Ex: bicycle frame hoist E 油壓缸 Engine jig.

  19. Moving part F CAN (2) Machine A structure composed of pin-connected multiforce members with moving parts is designed to transmit and alter the effect of forces. Ex:crusher 2. Assumptions (1) The structure is properly supported. (2) The structure contains no more supports or members than necessary to prevent its collapse. (3) The joint reactions of the structure can be determined from the equilibrium equations.

  20. 200N B C 2m 2m 3m 60 A 3. Analysis procedure (1) Draw the free body Diagram of entire structure, a part of structure, or each of its members. (a) Isolate each part by drawing its outline shape. Show all forces and/or couple moments act on the part. (b) Identify all the two-force member in the structure. (c) Forces common to any two contacting members act with equal magnitudes but opposite sense on respective members.

  21. 200N Cx 2m 60 2m Cy 0 60 A Free body diagrams of members AB & BC. By inspection, member AB is a two-force member. (2) Apply the equations of equilibrium (a) Count the total number of unknowns to make sure that an equivalent number of equilibrium equations can be written for solutions. , Cx , Cy (3) Unkoown:

  22. (b) Moments should be summed about a point that lies at the intersection of the lines of action of as many unknown forces as possible. Equations of equilibrium for member BC. (2) Check the correct sense of the unknown forces.

  23. 第六章 完

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