SM 1 05 Statics 5 Statically determined bar

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

SM 1 -05: Statics 5: Statically determined bar structures Trusses Formal definition: A frame is a plane (2 D) set of straight bars connected at hinged joints (corners) loaded at hinges by concentrated forces 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. Unstable! Frame M. Chrzanowski: Strength of Materials Truss 2/10

SM 1 -05: Statics 5: Statically determined bar structures Trusses Motivation to use trusses is quite different Trusses are aimed to span large areas with a light but durable structures Frame Truss M. Chrzanowski: Strength of Materials 3/10

SM 1 -05: Statics 5: Statically determined bar structures M. Chrzanowski: Strength of Materials Trusses 4/10

SM 1 -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 Tie The structure has to be kinematically stable! M. Chrzanowski: Strength of Materials 5/10

SM 1 -05: Statics 5: Statically determined bar structures Trusses Too many joints, too few bars! w – number of joints p – number of bars Too many bars, too few joints ν = 2 w – p – 3 Structure has to be kinematically stable, but can be statically determine or in-determine! Kinematics 2 w = number of equations p +3 = number of unknowns M. Chrzanowski: Strength of Materials Statics ν>0 ν=0 unstable undefined determine ν<0 stable In-determine 6/10

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

SM 1 -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. B A node can be cut off; then we have two equations of equilibrium (C) , or any bar with one equation of equilibrium (D). Y X = 0 A Y = 0 B MK = 0 M. Chrzanowski: Strength of Materials X = 0 C X = 0 Y = 0 X X D X = 0 Y = 0 8/10

SM 1 -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”. A A B C C 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 B. When a node connects two bars but the loading acts along of any of these bars C. When unloaded node connects three bars, two of them being co-linear M. Chrzanowski: Strength of Materials 9/10

SM 1 -05: Statics 5: Statically determined bar structures How does truss work? P=2 A 5 7 8 9 12 1 2 B 7 8 9 1 2 7 11 1 3 6 12 R=1 10 3 6 8 9 2 4 13 5 12 10 13 5 11 C All forces in k. N 6 11 R=1 Trusses 4 10 Bars under tension 4 Bars under compression „ 0 -bars” 13 3 M. Chrzanowski: Strength of Materials 10/10

SM 1 -05: Statics 5: Statically determined bar structures Trusses stop M. Chrzanowski: Strength of Materials 11/10