1. 1 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses STATICALLY DETERMINED PLANE BAR STURCTURES (TRUSSES)

2. 2 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses Frame The simplest truss consists of three bars connected in hinges. If the only loading will be forces acting at hinges then only cross sectional will be normal forces which can be found considering equilibrium of hinges. Formal definition: A frame is a plane (2D) set of straight bars connected at hinged joints (corners) loaded at hinges by concentrated forces Truss Unstable!

3. 3 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses Frame Motivation to use trusses is quite different Truss Trusses are aimed to span large areas with a light but durable structures

4. 4 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses

5. 5 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses Under the assumptions: if the structure consists of straight bars connected and loaded at hinges its elements have to bear the normal cross-sectional forces only Strut The structure has to be kinematically stable! Tie

6. 6 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses Structure has to be kinematically stable, but can be statically determine or in-determine! ν = 2w – p – 3 ν < 0 ν = 0 ν > 0 Kinematics stable unstable Statics determine In-determine undefined w – number of joints p – number of bars Too many joints, too few bars! Too many bars, too few joints 2w = number of equations p +3 = number of unknowns

7. 7 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses Examples of kinematically and statically determination Externally determined ν = 2·10 – 16 – 3 = 1 ν = 2w – p – 3 Internally indetermined w = 10, p =16 Kinematically stable Internally and externally determined ν = 2·10 – 17 – 3 = 0 w = 10, p = 17 ν = 2·10 – 17 – 4 = -1 w = 10, p = 17 Internally determined Externally indetermined Kinematically stable

8. 8 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses Normal forces in truss elements are to be found from the fundamental axiom that if the whole structure is in equilibrium then any part of it is in equilibrium, too. Such a part of a structure can be obtained by cutting off the truss through three bars not converging in a point (A), or through two bars converging in a node. (B). In the former case we have three equations of equilibrium, in the latter – two. A node can be cut off; then we have two equations of equilibrium (C), or A B X C X Y D  X = 0  Y = 0  M K = 0  X = 0  Y = 0  X = 0 any bar with one equation of equilibrium (D).  X = 0  Y = 0 B

9. 9 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses Certain bars are required solely for the purpose of keeping a truss kinematically stable. The cross-sectional forces in these bars vanish; one can call them „0-bars”. There are three cases in which we can easily spot „0-bars”: A. When only two bars converge in a node which is free of loading C. When unloaded node connects three bars, two of them being co-linear B. When a node connects two bars but the loading acts along of any of these bars B A C C A

10. 10 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses How does truss work? 12 5 34 78910 111213 A B C Bars under compression Bars under tension „0-bars” P=2 R=1 All forces in kN 12 5 34 6 7 89 10 11 12 13 12 5 34 6 7 8 910 11 12 13 6

11. 11 /10 M.Chrzanowski: Strength of Materials SM1-05: Statics 5:Statically determined bar structures Trusses  stop