User:Eml4500.f08.a-team.robinson/Lecture 9/19/08

Element 1: k (1) d (1) = f (1)

| 0.5625  0.32476  ... ...   |          |  0  |    |                             |          |  0  |    |                             |          |  0  |    |                             |4x4       |  0  |4x1

| -4.4348  |       |  f1(1)  | =  |  -2.5622  |  =    |  f2(1)  | | +4.4378  |       |  f3(1)  | | +2.5522  |       |  f4(1)  | f1(1) & f2(1) = reactions f3(1) & f4(1) = internal forces

Observation: Element 1 is in equilibrium summation Fx = f1(1) + f3(1) = 0 summation Fy = f2(1) + f4(1) = 0

p1(1) = [(f1(1))2 + f2(1))2]1/2

Two methods for solving 2-bar truss with Euler cut principle

Method 1:

Method 2:

Back to FE solution: to bring p back into the big picture --> equilibrium of global node n(circled)