# Tutorial 4a – FE-beam problem

In this tutorial we use Fesslix to analyze a finite element beam.

## 1   Mechanical model

We investigate the problem of the following linear elastic finite element beam:

Figure 1: Linear elastic beam investigated in this tutorial.

where length of beam , Young's modulus , stiffness of the rotational spring , point loads , and distributed load . The moment of inertia of the beam is .

## 2   Step by Step Instruction

### 2.1   Getting started

First of all, we need to load the finite element module of Fesslix:

Fesslix parameter file

Next, we define the parameter values of the mechanical model as constants:

Fesslix parameter file
const E = 7e10; #[N/m^2] Young's modulus
const I = 6.75e-4; #[m^4] moment of inertia
const K = 25e7; #[Nm/rad] stiffness of rotational spring
const F = 35e3; #[N] point load
const q = 1e4; #[Nm] distributed load
const l = 3; #[m] length of beam

### 2.2   Define the mesh

First, we need to define the nodes:

Fesslix parameter file
node 0 [0,0] fix wx, fix wy, fix wz, fix rx, fix ry, spring rz K;
node 1 [l/2,0] fix wx, fix wz, fix rx, fix ry;
node 2 [l,0] fix wx, fix wz, fix rx, fix ry;

Thereafter, we can connect the nodes using so-called edges:

Fesslix parameter file
edge 1 0 1; # Connect nodes 0 and 1
edge 2 1 2; # Connect nodes 1 and 2

### 2.3   Define the finite elements

Before the actual definition of the finite elements, we need to define the material and the cross-section to be employed:

Fesslix parameter file
# Material that has Young's modulus E
isomat 0 E E nu 0.0;

# Cross-section with moment of inertia I
crosssec 0 Manual A 0 Iy I Iz I;

Now we can define a FE-beam:

Fesslix parameter file
elbeam 1 1 { material=0; };
elbeam 2 2 { crosssec=0; };

Fesslix parameter file
#! –––––––––––––––-
#! –––––––––––––––-
# Apply force F at node 1 and assign it to load-case 0
# Apply force F at node 2 and assign it to load-case 0

#! –––––––––––––––-
#! –––––––––––––––-
# Apply distributed load q to the 2nd beam

### 2.5   Solve the FE-system

Fesslix parameter file
#! –––––––––––––––-
#! Assemble degrees of freedom
#! –––––––––––––––-
fem assdof { printmtx=true; };

#! –––––––––––––––-
#! Assemble the stiffness matrix
#! –––––––––––––––-
fem assstf { printmtx=true; };

#! –––––––––––––––-
#! Assemble the force vector
#! –––––––––––––––-
fem assforces 0,1 { printmtx=true; };
# ... employing load cases 0 and 1

#! –––––––––––––––-
#! Solve the problem
#! –––––––––––––––-
fem solve { };
#fem solve { pcn = 4; };
# the conjugate gradient method is used to solve the system
# employing a diagonal preconditioner.
# To solve the system using a direct solver, the optional parameter
# 'pcn' can be set to '4'.

### 2.6   Post-processing

To output the results at the degrees of freedom of the model:

Fesslix parameter file

fem print nodes { printsol=true; };

To print the vertical displacement at the nodes of the model:

Fesslix parameter file

funplot nodeu([1,wy]), nodeu([2,wy]);

This should return 4.689167e-3 and 0.01292 as result.

## 3   The complete input files of this tutorial

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