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2. Lecture 2 The Finite Element Analysis Process 2.092/2.093, Fall â09
Assume:
âą
frictionless joints
âą
weightless bars
âą
loads applied only at the joints
joints and elements cannot take moments truss element does not bend
âą â
âą concentrated loads only at the joints
static condition no vibrations/no transient response
âą â
âą infinitesimally small displacements, stress below yield stress
So, we perform a linear elastic analysis. Let A = area of each truss element, and E = Youngâs modulus. Each element has stiffness EA , where Li is the length of element i. The displacements of the nodes are
Li
defined as:
Equilibrium of the structure:
Y2(= R6)=2P , Y1(= R8)= âP , X1(= R7)=0
Typical element:
F1 and F2 have the same magnitude and are in the opposite direction, acting onto the element (resulting in tension in the element). 2
3. Lecture 2 The Finite Element Analysis Process 2.092/2.093, Fall â09
Equilibrium of each joint:
The result is âsystem equilibriumâ:
We see that every element and every joint is in equilibrium, including element forces and reactions and externally applied loads! Also, equilibrium holds for every part of the structure when we include all forces acting on that part. For example, this portion of the structure is in equilibrium: 3
4. Lecture 2
The Finite Element Analysis Process
2.092/2.093, Fall â09
Actual Solution
KU = R
(where K is the stiffness matrix)
â€âĄ
â€âĄ u1 R1
U =
u8 R8
This yields a complete model when the reactions are included. In this case, we have u6 = u7 = u8 = 0 and we use only the 5 Ă 5 matrix for K.
U =
âąâŁ âąâŁ â„⊠âąâŁ â„âŠ
. .
R =
. . ,. .
â€âĄ
â€âĄ
u1 R1
â„⊠âąâŁ â„âŠ
. .
R =
. . ,. .
u5 R5 4
5. MIT OpenCourseWare
http://ocw.mit.edu
2.092 / 2.093 Finite Element Analysis of Solids and Fluids I
Fall 2009
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