FINITE ELEMENT METHOD (INTRODUCTION) E, I, L F x Element 1 2 θ1 θ2 y1 y2 d y F y(x) θ(x)

Displacement vector of a single finite element having 4 DOF

C Potential energy

This integral can be calculated using symbolic operations in Matlab

Element Stiffness Matrix Example: 20 mm 2 mm 500 mm 1 N d=0.5 E=2x1011 Tip displacement is calculated using single element

y1=0 Sınır Şartı Sınır Şartı θ1=0 Force is applied at node 2 in positive y direction y1=0 Sınır Şartı θ1=0 Sınır Şartı

Recalculate using two elements Example: 20 mm 2 mm 500 mm 1 N 1 2 3 d=0.25 m x2 +6d +6d x2

Displacement and slope can be calculated at mid point of the beam y2=0.0049 m θ2=0.0354 rad y3=0.0157 m θ3=0.0471 rad Displacement and slope can be calculated at mid point of the beam Sınır şartı y1=0 θ1=0 Force is applied at node 3 in positive y direction

Analysis using SolidWorks Steel beam having 20x2x500 mm3 cross section Fixed 1 N δL=0.0155 m

Example: 1 N L=500 mm 2 mm δ 20 mm 1 2 3 Using two elements y1=0 Sınır şartı y1=0 y3=0 θ1=0.00584112149533 rad y2= 0.00097428081192 m θ2= -0.00000000000000 rad θ3= -0.00584112149533 rad Slope is zero at the mid point of the beam Slope has opposite sign at the ends of the beam Force is applied at node 2 in positive y direction