Linear Spring as a Finite Element and its Calculations

Share
 

Linear elastic spring is one of the mechanical device used in day to day life, which is capable of resisting axial loading only, and the contraction or elongation of the spring is directly proportional to the applied along the direction of the load. Proportionality constant between deformation and applied load is referred to as the spring constant(k), recoil rate, or spring stiffness k, and it has units of force per unit extent. As an elastic spring supports axial loading , we select an element coordinate system (which is also known as a local coordinate system) as an x-axis oriented along the spring length, as shown in the below figure.

a) Linear spring element with nodes, nodal displacements, and nodal forces.

linear spring equation

b) Load-deflection curve.

SLOPE

Let us assume that both the nodal displacements are zero when the spring is undeformed condition, the net spring deformation is given by

δ= U2− U1

Where,

δ” is net spring deformation

and the resultant axial force in the spring is

f = kδ= k(U2− U1)

Where,

f” is resultant axial force

For equilibrium condition,

f1+ f2= 0 (or) f1= − f2,

Then, in terms of the applied nodal forces as

f1= −k(U2− U1)

f2= k(U2− U1)

which can be expressed in matrix form as

 matrix{2}{2}{k -k -k k} *matrix{2}{1}{u1 u2}matrix{2}{1}{f1 f2}
or
[Ke]{U}={F}
Where,
[Ke] = matrix{2}{2}{k -k -k k}
{U} = matrix{2}{1}{u1 u2}
{F} = matrix{2}{1}{f1 f2}
[Ke] is “Stiffness Matrix for Spring Element” which is defined as the element stiffness matrix in the element coordinate system (or local system), where {u} is the column matrix (vector) of nodal displacements, { f } is the column matrix (vector) of element nodal forces.
matrix{2}{1}{f1 f2}= [Ke]*matrix{2}{1}{u1 u2}
with,
[Ke] = matrix{2}{2}{k -k -k k}
Lets take, {F} = [K] {X}, Where {F} is known, and [K] {X} is unknown.
Where,
{F} known term indicates, matrix{2}{1}{f1 f2}
 [K] indicates, [Ke]
{X} unknown term indicates, matrix{2}{1}{u1 u2}
Above equation shows that the element stiffness matrix for the linear spring element is a 2 * 2 matrix. Which corresponds to the fact that the element reveal two nodal displacements values ( or in other words it says degrees of freedom) that the two displacements are not independent values (which defines that, the body is continuous and elastic in nature).
Moreover, the matrix is symmetric. Which is a consequence of the symmetry of the forces ( that is equal and opposite to ensure equilibrium condition).
Here the matrix is singular and therefore the matrix is not invertible. That is why the problem is defined as incomplete one and it does not have a solution: which required a boundary conditions.
Hence, this is the derivation for “Linear Spring as a Finite Element and it calculations”.

Want to Study Aerospace Engineering?

Find out key answers in our checklist



  • Comments on Facebook

Write This!

Authors

Hello, I am an aircraft structural analyst with industrial experience and a master degree on aerospace structures. Currently working for an aerospace company as a stress
Advertisement
Top Aerospace Engineering Universities List Aerospace Engineering School's Videos
  • Most Views
  • Lastest
  • Comments
yorum ikonu
2016-11-09 13:58:00
yorum ikonu
2016-11-08 23:45:20
Subscribe to newsletter
Aviation Events

Aerospace Engineering

Aerospace Engineering and Aviation website provides information for universities, jobs, salary and museums for aeronautical, space and astronautical domain