University of Florida/Eml4500/f08.qwiki/Lecture 4

From Wikiversity
Jump to navigation Jump to search

Trusses, matrix method

Truss with 2 Elastic (deformable) Bars

Note: Points 1 and 3 are fixed (constrained) to zero in both the x and y directions

Global FBD

Note: There are 4 unknown reactions and three equations of equilibrium making this truss statically indeterminant

Seperate FBD's of the bar elements

Bar Element 1

Note: The superscript 1 in parenthesis on each force variable denotes that the force is applied to element 1

Bar Element 2

Note: The superscript 2 in parenthesis on each force variable denotes that the force is applied to element 2

Next Big Step:Force displacement relation (FD)

Recall: Spring 1.JPG

FD relation of 1-D spring element:

  • 1 end fixed: f = kd (see above figure)
  • 2 ends free:

Spring 2.JPG f2x1=k2x2d2x1

Case 1: Observer sits on node 1

  • f2 = k(d2-d1)

Case 2: Observer sits on node 2

  • f1+f2=0 Therefore, f1 = k(d1-d2)