Stiffness & Stiffness Matrix

Whenever we have a body and there is a load applied to it, we look for the deformation and need a relation between load and deformation.

Stiffness is easy to understand for the case of a spring loaded under a tensile force (F), where the end of the spring undergoes a displacement (u).

Stiffness (k) can be defined as            F = k * u,                     k = F/u

Therefore stiffness can be defined as the Force acting per unit displacement from the above equation.

Extending this concept to a bar with a tensile force (F) where end of the bar is undergoes a displacement (u).

Stiffness (k) = AE/L,

Where;            A = Cross sectional area of bar

E = Young’s Modulus

L = Length of the bar

Therefore;       F = (AE/L) * u

Now extending the concept to the case below,

Where;            P = Load on the bar

A1 & A2 = Cross sectional areas of the bar

We can clearly define the body as a combination of two bars with different stiffness’s. Once we have considered that we have said to have discretized the body.

This can also be represented as,

And further simplified as,

Considering the springs to be in equilibrium individual stiffness matrix for each spring is defined as,

Where the subscript is the element number.

Now we begin assembling the individual stiffness matrices of the elements considering that

Step 1

Step 2

Where,

Therefore,

Here,

 the system stiffness matrix.

Also k1 = A1E1/L1 & k2 = A2E2/l2 when using this stiffness matrix and the force displacement relation for the bar case.

Once we have written the Force-Displacement Relation we apply the boundary conditions for the above problem.

Boundary Conditions are: U1 =0, F2 = 0 & F3 = P

Now the equations that form would be:

Equation 1:     -k1*U2 = F1    This equation is called the constraint equation and F1 forms the unknown reaction acting at node 1.

 Gives Equation 2 & 3

Further simplified as

&

 Which give the values of displacement U2 & U3.

Next step would be to determine the Strain in the system using the Strain – Displacement relationship and then ultimately the stresses using the Stress – Strain relationship.

Strain-Displacement relationship would be given as

Having known the strain in element 1 & 2, using the Stress-Strain relationship.

σ1 = E1 * ϵ1 = E1 * (u2/L1)

u2 = P/k1, also k1= (A1*E1/L1) so u2 = P*L1/A1*E1

Therefore

σ1 = E1 * (1/L1) * (P*L1/A1*E1)

Therefore,

σ1 = P/A1

Similarly solving for (σ2 = E2 * ϵ2) gives,

σ2 = P/A2

This procedure is in a nutshell FEA or FEM, finite element analysis. To summarize we looked at a simple bar under tension and therefore we decided on the behavior of the bar under tensile loading. Next since it was made of two parts we divided the bar into two elements, i.e we discretized it and also defined K1 & K2 in the process.

Then we defined the stiffness matrix for the individual elements and then assembled the global stiffness matrix for the complete bar.

Then we applied an important aspect of finite elements i.e. boundary conditions. It is important to understand that we cannot solve a problem without applying the boundary conditions.

Further to which we solve and get the displacement of individual nodes. Then using the Strain – Displacement relationship we determine the strain in each element.

Finally we use the strains in the Stress – Strain relationship to calculate the stresses in each element.

This case we considered is the case of a bar with uniaxial load, where it was not difficult to understand stress. Stress would in this case be Young’s modulus multiplied by Strain. Real world engineering problems are not as straightforward as this case of one dimensional bar. All of them are 3D problems, where deformations are not homogeneous which means that stress varies from point to point.