7Zamonaviy ta'limda matematika, fizika va raqamli texnologiyalarning dolzarb muammolari va yutuqlari
Toshkent viloyati Chirchiq davlat pedagogika instituti
MODELING SPRING AS A FINITE ELEMENTS IN C++ BUILDER
O. T. Rajabov
Chirchik State Pedagogical Institute of Tashkent region, Chirchik, Uzbekistan
bryus 75@gmail.com
A liner elastic spring is a mechanical device capable of supporting axial loading only, and the elongation or contraction of the spring is directly proportional to the applied axial load. The constant of proportionality between deformation and load is referred to as the spring constant, spring rate, or spring stiffness k, and has units of force per unit length. As an elastic spring supports axial loading only, we select an element coordinate system (also known as a local coordinate system) as an x axis oriented along the length of spring, as shown.
Pic №1 Liner spring element with nodes, nodal displacements, and nodal forces.
Pic №2 Load-deflection curve Assuming that both the nodal displacement are zero when the spring is undeformed, the spring deformation is given by b=u2+u2 and the resultant axial force in the spring is: f=kS=k(u2+u2) For aquilibrium,
fi+f2=0 or fi=-f2 (1)
Then, in terms of the applied nodal forces as:
fi=-k(u2-ui) (2)
f2=k(u2-u1) (3)
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Academic Research, Uzbekistan 791 www.ares.uz
Zamonaviy ta'limda matematika, fizika va raqamli texnologiyalarning dolzarb muammolari va yutuqlari
Toshkent viloyati Chirchiq davlat pedagogika instituti
which can be expressed in matrix form as: k -k {fi L-k k J W I/2 -k k
or [ke][u} = {f}
(4)[1]
(Where
k -k
Stiffness matrix for one spring element is defined as the element
stiffness matrix in the element coordinate system (or local system), {u} is the column matrix (vector) of nodal displacements, and {/} is the column matrix (vector) of element nodal forces.
(4) equation shows that element stiffness matrix for linear spring element is 2x2 matrix. This corresponds to the fact that the element axhibits two nodal displacement (or degrees of freedom) and that the two displacement are not independent (that is, the body is continuous and elastic). Furthermore, the matrix is symmetric. This is a consequence of the symmetry of the forces (equal and opposite to ensure equilibrium). Also the matrix is singular and therefore not in invertible. That is because the problem as defined is incomplete and does not have a solution: boundry conditions are required. By (4) quation we find forces on each nodes by c++ and Borland c++ GUI
Pic №2- GUI interface We find as a result nodal displacement |u1-u2|=0.83-0.7=0.13
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Zamonaviv ta'limda matematika, fizika va raqamli texnologiyalarning dolzarb muammolari va vutuqlari
Toshkent viloyati Chirchiq davlat pedagogika instituti
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Zamonaviy ta'limda matematika, fizika va raqamli texnologiyalarning dolzarb muammolari va yutuqlari
Toshkent viloyati Chirchiq davlat pedagogika instituti
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