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

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

Separate 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:

FD relation of 1-D spring element:


 * 1 end fixed: f = kd (see above figure)

f 2x1= k 2x2 d 2x1
 * 2 ends free:

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)